Operations on Limits of Functions: Theorems, Rules and Indeterminate Forms

Hypotheses and the meaning of operations on limits

Operations on limits concern functions obtained by combining other functions pointwise. Let \(f\) and \(g\) be two real functions defined on a deleted neighbourhood of \(x_0\) — that is, on a neighbourhood of \(x_0\) from which the single point \(x_0\) has been removed — and suppose that \(x_0\) is an accumulation point of their common domain, a condition that is necessary for the limit to be meaningful. We may then consider, for instance, the functions

\[ (f+g)(x)=f(x)+g(x), \qquad (fg)(x)=f(x)g(x). \]

Likewise, at the points where \(g(x)\neq 0\), we may define

\[ \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}. \]

The problem is to decide whether the limit of the function so obtained can be determined from the limits of \(f\) and \(g\) considered separately. Suppose, to begin with, that

\[ \lim_{x\to x_0}f(x)=L \qquad\text{and}\qquad \lim_{x\to x_0}g(x)=M, \]

where \(L\) and \(M\) are real numbers. The theorems on operations with limits give the conditions under which the limit of the sum, the product, the quotient, or other combinations of the two functions may be recovered from \(L\) and \(M\).

It is important to understand that expressions such as

\[ \lim_{x\to x_0}\bigl(f(x)+g(x)\bigr) = \lim_{x\to x_0}f(x) + \lim_{x\to x_0}g(x) \]

are neither definitions nor purely formal manipulations. They express the conclusion of a theorem, and hold only when all its hypotheses are met.

Before applying a rule, one must also check that the function produced by the operation is in fact defined for values of \(x\) arbitrarily close to \(x_0\), with \(x\neq x_0\). In the case of a quotient, for example, it is not enough that \(f\) and \(g\) be defined: one must restrict attention to the points at which the denominator does not vanish.

Whatever value the functions may take at \(x_0\) plays no part in the computation. The limit depends on the behaviour of the functions near the point, not on the value they take at the point itself. For this reason the theorems remain valid even when one or both functions are undefined at \(x_0\).

Two fundamental limits

Before examining operations between functions, we recall two elementary results that will be used in the sections that follow.

If \(c\in\mathbb{R}\), then for the constant function \(f(x)=c\) we have

\[ \lim_{x\to x_0}c=c. \]

Indeed, for every \(\varepsilon>0\),

\[ |c-c|=0<\varepsilon, \]

regardless of the choice of \(x\). The same conclusion holds when the variable tends to infinity:

\[ \lim_{x\to+\infty}c=c, \qquad \lim_{x\to-\infty}c=c. \]

For the identity function \(f(x)=x\), on the other hand,

\[ \lim_{x\to x_0}x=x_0. \]

This relation follows at once from the definition, since

\[ |f(x)-x_0|=|x-x_0|. \]

At infinity, the identity function follows the behaviour of the variable:

\[ \lim_{x\to+\infty}x=+\infty, \qquad \lim_{x\to-\infty}x=-\infty. \]

Together with the theorems on operations, these limits will allow us to compute the limits of polynomials and of other algebraic expressions.

The first rules will be stated and proved for finite limits in the case \(x\to x_0\). The same rules, with the corresponding hypotheses, hold also for one-sided limits and for \(x\to+\infty\) or \(x\to-\infty\). The proofs are obtained by replacing the condition \(0<|x-x_0|<\delta\) with the appropriate one-sided condition, or with a condition of the form \(x>R\) or \(x<-R\). The cases in which a function tends to \(+\infty\) or \(-\infty\), by contrast, call for separate treatment, since some operations retain a definite meaning while others lead to indeterminate forms.

The limit of a sum and of a difference

Let \(f\) and \(g\) be two functions defined on a deleted neighbourhood of \(x_0\), and suppose that

\[ \lim_{x\to x_0}f(x)=L \qquad\text{and}\qquad \lim_{x\to x_0}g(x)=M, \]

with \(L,M\in\mathbb{R}\). Then the function \(f+g\) also has a limit as \(x\to x_0\), and

\[ \lim_{x\to x_0}\bigl(f(x)+g(x)\bigr)=L+M. \]

In other words, when both limits are finite, the limit of the sum equals the sum of the limits.

Proof. We must show that, for every \(\varepsilon>0\), there exists \(\delta>0\) such that

\[ 0<|x-x_0|<\delta \quad\Longrightarrow\quad \bigl|f(x)+g(x)-(L+M)\bigr|<\varepsilon. \]

We first observe that

\[ f(x)+g(x)-(L+M) = \bigl(f(x)-L\bigr)+\bigl(g(x)-M\bigr). \]

By the triangle inequality,

\[ \bigl|f(x)+g(x)-(L+M)\bigr| \leq |f(x)-L|+|g(x)-M|. \]

Since \(f(x)\to L\), there exists \(\delta_1>0\) such that

\[ 0<|x-x_0|<\delta_1 \quad\Longrightarrow\quad |f(x)-L|<\frac{\varepsilon}{2}. \]

Similarly, since \(g(x)\to M\), there exists \(\delta_2>0\) such that

\[ 0<|x-x_0|<\delta_2 \quad\Longrightarrow\quad |g(x)-M|<\frac{\varepsilon}{2}. \]

Put

\[ \delta=\min\{\delta_1,\delta_2\}. \]

If \(0<|x-x_0|<\delta\), both of the above estimates hold. Consequently,

\[ \begin{aligned} \bigl|f(x)+g(x)-(L+M)\bigr| &\leq |f(x)-L|+|g(x)-M|\\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\ &=\varepsilon. \end{aligned} \]

Therefore,

\[ \lim_{x\to x_0}\bigl(f(x)+g(x)\bigr)=L+M. \]

The limit of a difference

The difference may be rewritten as a sum:

\[ f(x)-g(x)=f(x)+\bigl(-g(x)\bigr). \]

Since

\[ \bigl|-g(x)-(-M)\bigr|=|g(x)-M|, \]

it follows from the definition of limit that

\[ \lim_{x\to x_0}\bigl(-g(x)\bigr)=-M. \]

Applying the theorem on sums, we obtain

\[ \lim_{x\to x_0}\bigl(f(x)-g(x)\bigr)=L-M. \]

We may therefore combine the two results in the single formula

\[ \lim_{x\to x_0}\bigl(f(x)\pm g(x)\bigr)=L\pm M. \]

The sum of finitely many functions

By repeated application of the theorem, the result extends to any finite number of functions. If

\[ \lim_{x\to x_0}f_k(x)=L_k \qquad \text{for }k=1,\ldots,n, \]

then

\[ \lim_{x\to x_0}\sum_{k=1}^{n}f_k(x) = \sum_{k=1}^{n}L_k. \]

The theorem concerns finite limits. When one or both functions tend to \(+\infty\) or \(-\infty\), the behaviour of the sum or difference must be examined separately.

The limit of a product

Let \(f\) and \(g\) be two functions defined on a deleted neighbourhood of \(x_0\), and suppose that

\[ \lim_{x\to x_0}f(x)=L \qquad\text{and}\qquad \lim_{x\to x_0}g(x)=M, \]

with \(L,M\in\mathbb{R}\). Then the product \(fg\) also has a limit as \(x\to x_0\), and

\[ \lim_{x\to x_0}f(x)g(x)=LM. \]

Before proving the theorem, we recall a fundamental consequence of the existence of a finite limit.

Local boundedness

If

\[ \lim_{x\to x_0}f(x)=L\in\mathbb{R}, \]

then \(f\) is bounded on a suitable deleted neighbourhood of \(x_0\). Indeed, taking \(\varepsilon=1\), there exists \(\delta_0>0\) such that

\[ 0<|x-x_0|<\delta_0 \quad\Longrightarrow\quad |f(x)-L|<1. \]

From the triangle inequality it follows that

\[ |f(x)| \leq |f(x)-L|+|L| < |L|+1. \]

Hence, sufficiently close to \(x_0\), the values of \(f(x)\) cannot become arbitrarily large in absolute value.

Proof of the theorem

Consider the difference between the product \(f(x)g(x)\) and the candidate limit \(LM\). Adding and subtracting \(f(x)M\), we obtain

\[ \begin{aligned} f(x)g(x)-LM &=f(x)g(x)-f(x)M+f(x)M-LM\\ &=f(x)\bigl(g(x)-M\bigr)+M\bigl(f(x)-L\bigr). \end{aligned} \]

Applying the triangle inequality,

\[ |f(x)g(x)-LM| \leq |f(x)|\,|g(x)-M| + |M|\,|f(x)-L|. \]

Since \(f(x)\to L\), local boundedness provides a number \(K>0\) and a number \(\delta_0>0\) such that

\[ 0<|x-x_0|<\delta_0 \quad\Longrightarrow\quad |f(x)|<K. \]

For instance, we may take \(K=|L|+1\).

Given \(\varepsilon>0\), since \(g(x)\to M\), there exists \(\delta_1>0\) such that

\[ 0<|x-x_0|<\delta_1 \quad\Longrightarrow\quad |g(x)-M|<\frac{\varepsilon}{2K}. \]

Moreover, since \(f(x)\to L\), there exists \(\delta_2>0\) such that

\[ 0<|x-x_0|<\delta_2 \quad\Longrightarrow\quad |f(x)-L|<\frac{\varepsilon}{2(|M|+1)}. \]

Put

\[ \delta=\min\{\delta_0,\delta_1,\delta_2\}. \]

If \(0<|x-x_0|<\delta\), then

\[ \begin{aligned} |f(x)g(x)-LM| &\leq |f(x)|\,|g(x)-M|+|M|\,|f(x)-L|\\ &<K\frac{\varepsilon}{2K} +|M|\frac{\varepsilon}{2(|M|+1)}\\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\ &=\varepsilon. \end{aligned} \]

We have thus shown that

\[ \lim_{x\to x_0}f(x)g(x)=LM. \]

Multiplication by a constant

If \(c\in\mathbb{R}\) is a constant and

\[ \lim_{x\to x_0}f(x)=L, \]

then, taking the constant function \(g(x)=c\), the product theorem gives

\[ \lim_{x\to x_0}c\,f(x)=cL. \]

This result allows any constant factor to be taken outside the limit sign.

The product of finitely many functions

By repeated application of the theorem, if

\[ \lim_{x\to x_0}f_k(x)=L_k \qquad \text{for }k=1,\ldots,n, \]

then

\[ \lim_{x\to x_0}\prod_{k=1}^{n}f_k(x) = \prod_{k=1}^{n}L_k. \]

The limit of a reciprocal and of a quotient

Passing to the reciprocal requires an additional condition: the limit of the function in the denominator must be non-zero. This hypothesis guarantees that the function does not vanish on a suitable deleted neighbourhood of the point in question.

The limit of a reciprocal

Let \(g\) be a function defined on a deleted neighbourhood of \(x_0\), and suppose that

\[ \lim_{x\to x_0}g(x)=M, \qquad M\neq 0. \]

Then there is a deleted neighbourhood of \(x_0\) on which \(g(x)\neq 0\), and

\[ \lim_{x\to x_0}\frac{1}{g(x)}=\frac{1}{M}. \]

The denominator stays away from zero

Since \(M\neq 0\), we may take

\[ \varepsilon_0=\frac{|M|}{2}. \]

From the convergence of \(g(x)\) to \(M\), there exists \(\delta_0>0\) such that

\[ 0<|x-x_0|<\delta_0 \quad\Longrightarrow\quad |g(x)-M|<\frac{|M|}{2}. \]

By the reverse triangle inequality,

\[ |g(x)| \geq |M|-|g(x)-M| > |M|-\frac{|M|}{2} = \frac{|M|}{2}. \]

Hence \(g(x)\neq 0\) for every \(x\) sufficiently close to \(x_0\), with \(x\neq x_0\). Moreover, the absolute value of the denominator is bounded below by a positive constant.

Proof. Consider the difference

\[ \frac{1}{g(x)}-\frac{1}{M}. \]

Combining over a common denominator, we obtain

\[ \left|\frac{1}{g(x)}-\frac{1}{M}\right| = \frac{|M-g(x)|}{|M|\,|g(x)|}. \]

For \(0<|x-x_0|<\delta_0\) we have \(|g(x)|>\displaystyle\frac{|M|}{2}\), and therefore

\[ \left|\frac{1}{g(x)}-\frac{1}{M}\right| < \frac{2}{|M|^2}|g(x)-M|. \]

Given \(\varepsilon>0\), since \(g(x)\to M\), there exists \(\delta_1>0\) such that

\[ 0<|x-x_0|<\delta_1 \quad\Longrightarrow\quad |g(x)-M|<\frac{\varepsilon |M|^2}{2}. \]

Setting

\[ \delta=\min\{\delta_0,\delta_1\}, \]

for \(0<|x-x_0|<\delta\) it follows that

\[ \left|\frac{1}{g(x)}-\frac{1}{M}\right| < \frac{2}{|M|^2}\cdot\frac{\varepsilon |M|^2}{2} = \varepsilon. \]

We have therefore shown that

\[ \lim_{x\to x_0}\frac{1}{g(x)}=\frac{1}{M}. \]

The limit of a quotient

Now let \(f\) and \(g\) be two functions defined on a deleted neighbourhood of \(x_0\), and suppose that

\[ \lim_{x\to x_0}f(x)=L, \qquad \lim_{x\to x_0}g(x)=M, \]

with \(L,M\in\mathbb{R}\) and \(M\neq 0\). Since

\[ \frac{f(x)}{g(x)} = f(x)\frac{1}{g(x)}, \]

applying the product theorem together with the reciprocal theorem, we obtain

\[ \lim_{x\to x_0}\frac{f(x)}{g(x)} = \frac{L}{M}. \]

The condition \(M\neq 0\) is no mere technicality. If the denominator tends to zero, the quotient theorem cannot be applied, and the limits of \(f\) and \(g\) are not, in general, sufficient to determine the result. In particular, when the numerator tends to zero as well, one meets the indeterminate form \(\displaystyle\frac{0}{0}\), which will be studied in the section devoted to it.

The limit of powers

The product theorem enables us to determine the limit of the integer powers of a function. Let \(f\) be defined on a deleted neighbourhood of \(x_0\), and suppose that

\[ \lim_{x\to x_0}f(x)=L, \qquad L\in\mathbb{R}. \]

Positive integer exponent

If \(n\in\mathbb{N}\) and \(n\geq 1\), then

\[ \lim_{x\to x_0}\bigl(f(x)\bigr)^n=L^n. \]

Indeed, the power \(\bigl(f(x)\bigr)^n\) is the product of \(n\) factors each equal to \(f(x)\):

\[ \bigl(f(x)\bigr)^n = \underbrace{f(x)\cdot f(x)\cdots f(x)}_{n\text{ factors}}. \]

By repeated application of the product theorem, we obtain

\[ \lim_{x\to x_0}\bigl(f(x)\bigr)^n = \underbrace{L\cdot L\cdots L}_{n\text{ factors}} = L^n. \]

Zero exponent

At the points where \(f(x)\neq 0\), we have

\[ \bigl(f(x)\bigr)^0=1. \]

If \(L\neq 0\), the function \(f\) remains non-zero on a suitable deleted neighbourhood of \(x_0\). On that neighbourhood the zeroth power is therefore defined, and

\[ \lim_{x\to x_0}\bigl(f(x)\bigr)^0=1=L^0. \]

If, on the other hand, \(L=0\), the expression \(L^0\) is undefined and the preceding formula cannot be used. Nevertheless, at the points where \(f(x)\neq 0\) one still has

\[ \bigl(f(x)\bigr)^0=1. \]

Hence, if \(x_0\) is an accumulation point of the domain of the function \(\bigl(f(x)\bigr)^0\), the corresponding limit equals \(1\).

Negative integer exponent

Let \(n\in\mathbb{N}\) with \(n\geq 1\). If \(L\neq 0\), then \(f(x)\neq 0\) for every \(x\) sufficiently close to \(x_0\), with \(x\neq x_0\), and we may write

\[ \bigl(f(x)\bigr)^{-n} = \frac{1}{\bigl(f(x)\bigr)^n}. \]

From the limit of positive powers together with the reciprocal theorem it follows that

\[ \lim_{x\to x_0}\bigl(f(x)\bigr)^{-n} = \frac{1}{L^n} = L^{-n}. \]

We may therefore summarise the results, distinguishing the several cases. For every positive integer \(n\),

\[ \lim_{x\to x_0}\bigl(f(x)\bigr)^n=L^n. \]

For \(n=0\), the function \(\bigl(f(x)\bigr)^0\) equals \(1\) at the points where \(f(x)\neq 0\); hence, if \(x_0\) is an accumulation point of its domain, the limit equals \(1\).

For \(n<0\), one requires \(L\neq 0\); in that case,

\[ \lim_{x\to x_0}\bigl(f(x)\bigr)^n=L^n. \]

Polynomials in a function

Combining the limit of powers with the theorems on sums and on multiplication by a constant, we obtain a more general result. If

\[ P(t)=a_0+a_1t+\cdots+a_nt^n \]

is a polynomial, then

\[ \lim_{x\to x_0}P\bigl(f(x)\bigr)=P(L). \]

Explicitly,

\[ \lim_{x\to x_0} \left( a_0+a_1f(x)+\cdots+a_n\bigl(f(x)\bigr)^n \right) = a_0+a_1L+\cdots+a_nL^n. \]

This shows that, when the limit of \(f\) is finite, a polynomial in \(f(x)\) preserves its algebraic structure under passage to the limit.

The limit of roots

Let \(n\in\mathbb{N}\), with \(n\geq 2\), and suppose that

\[ \lim_{x\to x_0}f(x)=L, \qquad L\in\mathbb{R}. \]

To determine the limit of \(\sqrt[n]{f(x)}\), we must distinguish the case of odd \(n\) from that of even \(n\).

Roots of odd index

If \(n\) is odd, the \(n\)th root is defined for every real number. Accordingly,

\[ \lim_{x\to x_0}\sqrt[n]{f(x)}=\sqrt[n]{L}. \]

The result holds for every real value of \(L\), whether positive, zero, or negative.

Roots of even index

If \(n\) is even, the real \(n\)th root \(\sqrt[n]{f(x)}\) is defined only at the points where \(f(x)\geq 0\). Consequently, the limit must be taken within the domain of the function.

If \(L>0\), the convergence of \(f(x)\) to \(L\) implies that \(f(x)>0\) for every \(x\) sufficiently close to \(x_0\). In this case,

\[ \lim_{x\to x_0}\sqrt[n]{f(x)}=\sqrt[n]{L}. \]

The same formula holds when \(L=0\), provided \(x_0\) is an accumulation point of the domain of the function \(x\mapsto\sqrt[n]{f(x)}\):

\[ \lim_{x\to x_0}\sqrt[n]{f(x)}=0. \]

If, on the other hand, \(L<0\), then by preservation of sign we have \(f(x)<0\) for every \(x\) sufficiently close to \(x_0\), with \(x\neq x_0\). The even-index root is therefore undefined over the reals on any sufficiently small deleted neighbourhood of \(x_0\), which cannot then be an accumulation point of its domain.

Proof. Put

\[ u(x)=\sqrt[n]{f(x)} \qquad\text{and}\qquad a=\sqrt[n]{L}, \]

in the cases where these quantities are real.

If \(L=0\), then \(a=0\), and, at the points belonging to the domain of the root,

\[ |u(x)-a| = \left|\sqrt[n]{f(x)}\right| = \sqrt[n]{|f(x)|}. \]

Given \(\varepsilon>0\), the convergence of \(f(x)\) to \(0\) yields \(\delta>0\) such that

\[ 0<|x-x_0|<\delta \quad\Longrightarrow\quad |f(x)|<\varepsilon^n. \]

It follows that

\[ |u(x)| = \sqrt[n]{|f(x)|} < \varepsilon, \]

and hence \(u(x)\to 0\).

Now suppose \(L\neq 0\). If \(n\) is even, then necessarily \(L>0\), and hence \(a>0\). Given \(\varepsilon>0\), put

\[ \varepsilon_0= \begin{cases} \varepsilon, & n\text{ odd},\\ \min\left\{\varepsilon,\displaystyle\frac{a}{2}\right\}, & n\text{ even}. \end{cases} \]

The function \(t\mapsto t^n\) is strictly increasing on \(\mathbb{R}\) when \(n\) is odd, and on \([0,+\infty)\) when \(n\) is even. The numbers

\[ \eta_1=(a+\varepsilon_0)^n-a^n, \qquad \eta_2=a^n-(a-\varepsilon_0)^n \]

are therefore both positive. Put

\[ \eta=\min\{\eta_1,\eta_2\}. \]

Since \(f(x)\to L=a^n\), there exists \(\delta>0\) such that

\[ 0<|x-x_0|<\delta \quad\Longrightarrow\quad |f(x)-a^n|<\eta. \]

At the points belonging to the domain of the root, this inequality implies

\[ (a-\varepsilon_0)^n < f(x) < (a+\varepsilon_0)^n. \]

By the monotonicity of the \(n\)th root it follows that

\[ a-\varepsilon_0 < u(x) < a+\varepsilon_0. \]

Consequently,

\[ |u(x)-a|<\varepsilon_0\leq\varepsilon. \]

We have thus shown that

\[ \lim_{x\to x_0}\sqrt[n]{f(x)}=\sqrt[n]{L} \]

provided \(x_0\) is an accumulation point of the real domain of the function \(x\mapsto\sqrt[n]{f(x)}\).

Powers with rational exponent

Let \(m\in\mathbb{Z}\) and \(n\in\mathbb{N}\), with \(n\geq 1\), and suppose the fraction \(\displaystyle\frac{m}{n}\) is in lowest terms. If \(x_0\) is an accumulation point of the real domain of the function

\[ x\mapsto\bigl(f(x)\bigr)^{\frac{m}{n}} \]

and the number \(L^{\frac{m}{n}}\) is defined over the reals, then

\[ \lim_{x\to x_0}\bigl(f(x)\bigr)^{\frac{m}{n}}=L^{\frac{m}{n}}. \]

The result is obtained by composing the limit of integer powers with that of roots: one writes \(\bigl(f(x)\bigr)^{\frac{m}{n}}=\sqrt[n]{\bigl(f(x)\bigr)^{m}}\) and applies in turn the two theorems already proved.

If \(n\) is even, the base must be non-negative at the points considered; when \(m<0\), it must in addition be strictly positive, and one requires \(L>0\). If \(n\) is odd, negative bases are also admitted; when \(m<0\), the base must nonetheless be non-zero at the points considered, and \(L\neq 0\).

The limit of an absolute value

Let \(f\) be a function defined on a deleted neighbourhood of \(x_0\), and suppose that

\[ \lim_{x\to x_0}f(x)=L, \qquad L\in\mathbb{R}. \]

Then

\[ \lim_{x\to x_0}|f(x)|=|L|. \]

Proof. By the reverse triangle inequality, for every \(x\) in the domain of \(f\) we have

\[ \bigl||f(x)|-|L|\bigr| \leq |f(x)-L|. \]

Given \(\varepsilon>0\), the convergence of \(f(x)\) to \(L\) yields \(\delta>0\) such that

\[ 0<|x-x_0|<\delta \quad\Longrightarrow\quad |f(x)-L|<\varepsilon. \]

Under the same conditions we then have

\[ \bigl||f(x)|-|L|\bigr| \leq |f(x)-L| < \varepsilon. \]

This proves that

\[ \lim_{x\to x_0}|f(x)|=|L|. \]

The converse fails in general

The existence of the limit of \(|f(x)|\) does not entail the existence of the limit of \(f(x)\). Consider, for example,

\[ f(x)= \begin{cases} 1, & x>0,\\ -1, & x<0. \end{cases} \]

For every \(x\neq 0\) we have \(|f(x)|=1\), so that

\[ \lim_{x\to 0}|f(x)|=1. \]

However,

\[ \lim_{x\to 0^+}f(x)=1 \qquad\text{and}\qquad \lim_{x\to 0^-}f(x)=-1. \]

Since the right-hand and left-hand limits differ, the limit of \(f\) as \(x\to 0\) does not exist.

Preservation of sign

If \(L>0\), then \(f(x)>0\) for every \(x\) sufficiently close to \(x_0\), with \(x\neq x_0\). Indeed, taking \(\varepsilon=L/2\), we obtain

\[ |f(x)-L|<\frac{L}{2}, \]

whence

\[ f(x)>L-\frac{L}{2}=\frac{L}{2}>0. \]

Similarly, if \(L<0\), the function takes negative values on a suitable deleted neighbourhood of \(x_0\).

When the limit is non-zero, the absolute value can therefore be removed locally, according to the sign of \(L\):

\[ |f(x)|= \begin{cases} f(x), & L>0,\\ -f(x), & L<0, \end{cases} \]

for every \(x\) sufficiently close to \(x_0\).

The limit of a composite function

Let \(f\) be a function defined on a deleted neighbourhood of \(x_0\), and let \(\varphi\) be a function defined on a deleted neighbourhood of a point \(L\). Suppose that

\[ \lim_{x\to x_0}f(x)=L. \]

The composite function

\[ (\varphi\circ f)(x)=\varphi\bigl(f(x)\bigr) \]

is defined at the points where the value \(f(x)\) belongs to the domain of \(\varphi\). To determine its limit, we must therefore study the behaviour of the outer function \(\varphi\) as its argument tends to \(L\).

Theorem on the limit of a composite function

Suppose that

\[ \lim_{t\to L}\varphi(t)=M, \]

with \(M\in\mathbb{R}\), and that, for every \(x\) sufficiently close to \(x_0\), with \(x\neq x_0\), the following conditions hold:

\[ f(x)\in\operatorname{Dom}(\varphi) \qquad\text{and}\qquad f(x)\neq L. \]

Then

\[ \lim_{x\to x_0}\varphi\bigl(f(x)\bigr)=M. \]

The condition \(f(x)\neq L\) is needed because the limit of \(\varphi(t)\) as \(t\to L\) describes the behaviour of the function at points near \(L\) but distinct from \(L\). The value \(\varphi(L)\), when it exists, plays no part in this formulation of the theorem.

Proof. Given \(\varepsilon>0\), from the relation

\[ \lim_{t\to L}\varphi(t)=M \]

it follows that there exists \(\eta>0\) such that

\[ 0<|t-L|<\eta \quad\Longrightarrow\quad |\varphi(t)-M|<\varepsilon. \]

Since \(f(x)\to L\), there exists \(\delta>0\) such that

\[ 0<|x-x_0|<\delta \quad\Longrightarrow\quad |f(x)-L|<\eta. \]

By hypothesis, shrinking \(\delta\) if necessary, we may further assume that

\[ f(x)\in\operatorname{Dom}(\varphi) \qquad\text{and}\qquad f(x)\neq L. \]

Therefore,

\[ 0<|f(x)-L|<\eta, \]

and, invoking the condition on the limit of \(\varphi\), we obtain

\[ \bigl|\varphi(f(x))-M\bigr|<\varepsilon. \]

It follows that

\[ \lim_{x\to x_0}\varphi\bigl(f(x)\bigr)=M. \]

The case of a continuous outer function

If \(\varphi\) is continuous at the point \(L\), then

\[ \lim_{t\to L}\varphi(t)=\varphi(L). \]

In this case there is no need to require \(f(x)\neq L\) near \(x_0\), for the outer function is defined at \(L\) as well, and its value there coincides with the limit. We thus obtain the rule

\[ \lim_{x\to x_0}\varphi\bigl(f(x)\bigr) = \varphi\left(\lim_{x\to x_0}f(x)\right) = \varphi(L). \]

This formula expresses the fact that the limit may be carried through a continuous function.

For instance, since the cosine function is continuous at every real point, if \(f(x)\to L\), then

\[ \lim_{x\to x_0}\cos\bigl(f(x)\bigr)=\cos L. \]

The role of the domain

Before applying the theorem, one must always check that the composition is defined at the points considered. If, for example, the outer function is

\[ \varphi(t)=\sqrt{t}, \]

then we need \(f(x)\geq 0\) at the points of the domain sufficiently close to \(x_0\). Likewise, if

\[ \varphi(t)=\frac{1}{t}, \]

we require \(f(x)\neq 0\).

The theorem on composite functions therefore does not remove the domain restrictions imposed by the operations involved: it determines the limit only after those restrictions have been checked.

Operations with infinite limits

When a function tends to \(+\infty\) or \(-\infty\), the limit is not a real number. Expressions involving the symbols of infinity must therefore not be read as ordinary arithmetic equalities: they abbreviate theorems about the behaviour of functions.

In this section we consider only those combinations that determine the result uniquely. The operations for which the limits of the individual functions do not suffice will be examined separately.

Sum and difference

Let \(f\) and \(g\) be defined on a deleted neighbourhood of \(x_0\). If

\[ \lim_{x\to x_0}f(x)=+\infty \qquad\text{and}\qquad \lim_{x\to x_0}g(x)=M, \]

with \(M\in\mathbb{R}\), then

\[ \lim_{x\to x_0}\bigl(f(x)+g(x)\bigr)=+\infty. \]

Indeed, since \(g(x)\to M\), for \(x\) sufficiently close to \(x_0\) we have

\[ g(x)>M-1. \]

Given a real number \(A\), the relation \(f(x)\to+\infty\) implies that, eventually,

\[ f(x)>A-M+1. \]

Hence,

\[ f(x)+g(x)>A-M+1+M-1=A. \]

The sum thus exceeds any prescribed real threshold. In the same way,

\[ \begin{aligned} +\infty+M&=+\infty,\\ -\infty+M&=-\infty, \end{aligned} \qquad M\in\mathbb{R}, \]

where these expressions stand, in abbreviated form, for the corresponding results on limits.

If both functions tend to the same infinity, their effects reinforce one another:

\[ \begin{aligned} +\infty+(+\infty)&=+\infty,\\ -\infty+(-\infty)&=-\infty. \end{aligned} \]

For instance, in the first case, given \(A\in\mathbb{R}\), it suffices that

\[ f(x)>\frac{A}{2} \qquad\text{and}\qquad g(x)>\frac{A}{2} \]

in order that \(f(x)+g(x)>A\).

The difference is reduced to a sum by means of the identity

\[ f(x)-g(x)=f(x)+\bigl(-g(x)\bigr). \]

Among the determinate cases that follow are

\[ \begin{aligned} +\infty-M&=+\infty,\\ -\infty-M&=-\infty,\\ M-(+\infty)&=-\infty,\\ M-(-\infty)&=+\infty, \end{aligned} \qquad M\in\mathbb{R}. \]

If the two terms tend to infinities of opposite sign, we have in addition the determinate cases

\[ +\infty-(-\infty)=+\infty, \qquad -\infty-(+\infty)=-\infty. \]

A difference between two quantities that both tend to the same infinity, on the other hand, does not fall under these rules.

Product

Suppose that

\[ \lim_{x\to x_0}f(x)=+\infty \qquad\text{and}\qquad \lim_{x\to x_0}g(x)=M, \]

with \(M>0\). Since \(g(x)\to M\), eventually

\[ g(x)>\frac{M}{2}>0. \]

Given \(A>0\), the relation \(f(x)\to+\infty\) implies that, for \(x\) sufficiently close to \(x_0\),

\[ f(x)>\frac{2A}{M}. \]

Consequently,

\[ f(x)g(x)>\frac{2A}{M}\cdot\frac{M}{2}=A, \]

and therefore

\[ \lim_{x\to x_0}f(x)g(x)=+\infty. \]

The sign of the product depends on the signs eventually taken by the two factors. If \(M\neq 0\), the various cases may be summarised as follows:

\[ \begin{array}{c|cc} & M>0 & M<0\\ \hline f(x)\to+\infty & f(x)g(x)\to+\infty & f(x)g(x)\to-\infty\\ f(x)\to-\infty & f(x)g(x)\to-\infty & f(x)g(x)\to+\infty \end{array} \]

If both factors have infinite limits, the usual sign rules apply:

\[ \begin{aligned} (+\infty)(+\infty)&=+\infty,\\ (-\infty)(-\infty)&=+\infty,\\ (+\infty)(-\infty)&=-\infty,\\ (-\infty)(+\infty)&=-\infty. \end{aligned} \]

The product of a function tending to zero and a function tending to infinity in absolute value, by contrast, is not determined by this information alone.

Reciprocal and quotient

If

\[ \lim_{x\to x_0}g(x)=+\infty \qquad\text{or}\qquad \lim_{x\to x_0}g(x)=-\infty, \]

then

\[ \lim_{x\to x_0}\frac{1}{g(x)}=0. \]

Indeed, given \(\varepsilon>0\), the absolute value of \(g(x)\) eventually exceeds \(\displaystyle\frac{1}{\varepsilon}\). Hence,

\[ \left|\frac{1}{g(x)}\right| = \frac{1}{|g(x)|} < \varepsilon. \]

If \(f(x)\to L\in\mathbb{R}\), then applying the product theorem for finite limits to

\[ \frac{f(x)}{g(x)} = f(x)\frac{1}{g(x)}, \]

and recalling that \(\displaystyle\frac{1}{g(x)}\to 0\), we obtain

\[ \lim_{x\to x_0}\frac{f(x)}{g(x)}=0 \]

whenever the numerator has a finite limit and the denominator tends to \(+\infty\) or \(-\infty\).

If instead the numerator tends to \(+\infty\) or \(-\infty\) and the denominator tends to a real number \(M\neq 0\), the quotient reduces to the product of the numerator with a function tending to \(\displaystyle\frac{1}{M}\). The sign of the result then depends on whether the numerator tends to \(+\infty\) or \(-\infty\), and on the sign of \(M\).

Denominator tending to zero

To describe the behaviour of the reciprocal when the denominator tends to zero, one must know the sign with which it approaches zero.

If

\[ g(x)\to 0^+, \]

that is, \(g(x)\to 0\) through positive values near \(x_0\), then

\[ \frac{1}{g(x)}\to+\infty. \]

If, on the other hand,

\[ g(x)\to 0^-, \]

then

\[ \frac{1}{g(x)}\to-\infty. \]

Consequently, if \(f(x)\to L\neq 0\), the limit of the quotient depends on the sign of \(L\) and on the sign with which \(g(x)\) tends to zero:

\[ \begin{array}{c|cc} & g(x)\to 0^+ & g(x)\to 0^-\\ \hline L>0 & \displaystyle\frac{f(x)}{g(x)}\to+\infty & \displaystyle\frac{f(x)}{g(x)}\to-\infty\\ L<0 & \displaystyle\frac{f(x)}{g(x)}\to-\infty & \displaystyle\frac{f(x)}{g(x)}\to+\infty \end{array} \]

If the denominator tends to zero while changing sign arbitrarily close to \(x_0\), this information does not, in general, determine a single limit.

Powers and roots

For every positive integer \(n\), if \(f(x)\to+\infty\), then

\[ \bigl(f(x)\bigr)^n\to+\infty. \]

If instead \(f(x)\to-\infty\), the result depends on the parity of the exponent:

\[ \bigl(f(x)\bigr)^n\to \begin{cases} +\infty, & n\text{ even},\\ -\infty, & n\text{ odd}. \end{cases} \]

For \(n\geq 1\), the properties of the reciprocal further give

\[ \bigl(f(x)\bigr)^{-n}\to 0 \]

when \(f(x)\to+\infty\) or \(f(x)\to-\infty\).

Finally, if \(f(x)\to+\infty\), then

\[ \sqrt[n]{f(x)}\to+\infty \]

for every \(n\geq 2\). If \(n\) is odd and \(f(x)\to-\infty\), then instead

\[ \sqrt[n]{f(x)}\to-\infty. \]

All the rules presented in this section concern cases in which the limits of the functions involved determine the result. When this is not so, the operation leads to an indeterminate form.

Determinate and indeterminate forms

When one computes the limit of an expression built from several functions, the limits of the individual parts may either suffice to settle the result or leave several possibilities open.

In the first case one speaks of a determinate form: the behaviour of the functions involved fixes the limit of the expression uniquely. In the second case one speaks of an indeterminate form: the limits of the individual parts alone do not yield the result.

An indeterminate form does not therefore mean that the limit fails to exist. It means only that the limit cannot be found by direct application of the general rules on operations.

The meaning of the symbolic forms

Expressions such as

\[ \frac{0}{0}, \qquad \frac{\infty}{\infty}, \qquad 0\cdot\infty, \qquad \infty-\infty \]

do not denote operations carried out on numbers. They describe, in condensed form, the behaviour of the functions that make up the expression whose limit is being considered.

For example, the form

\[ \frac{0}{0} \]

means that the numerator tends to zero while, at the same time, the denominator tends to zero. From this information alone no definite value for the quotient follows.

The indeterminate form \(\displaystyle\frac{0}{0}\)

Consider the limits

\[ \lim_{x\to 0}\frac{x}{x}, \qquad \lim_{x\to 0}\frac{x^2}{x}, \qquad \lim_{x\to 0}\frac{x}{x^2}. \]

In all three cases the numerator and the denominator tend to zero. Nevertheless,

\[ \frac{x}{x}=1, \qquad \frac{x^2}{x}=x, \qquad \frac{x}{x^2}=\frac{1}{x}, \]

for \(x\neq 0\). Hence the first limit equals \(1\) and the second equals \(0\). In the third case,

\[ \lim_{x\to 0^+}\frac{1}{x}=+\infty \qquad\text{and}\qquad \lim_{x\to 0^-}\frac{1}{x}=-\infty. \]

Since the right-hand and left-hand limits differ, the limit as \(x\to 0\) does not exist.

The form \(\displaystyle\frac{0}{0}\) may therefore lead to a non-zero real number, to zero, to an infinite limit, or to the non-existence of the limit.

The indeterminate form \(\displaystyle\frac{\infty}{\infty}\)

Consider now

\[ \lim_{x\to+\infty}\frac{x}{x}, \qquad \lim_{x\to+\infty}\frac{x}{x^2}, \qquad \lim_{x\to+\infty}\frac{x^2}{x}. \]

In each quotient the numerator and the denominator tend to \(+\infty\), but

\[ \frac{x}{x}=1, \qquad \frac{x}{x^2}=\frac{1}{x}, \qquad \frac{x^2}{x}=x. \]

The respective limits are therefore

\[ 1, \qquad 0, \qquad +\infty. \]

The result depends on how fast the numerator and the denominator grow, not merely on the fact that both tend to infinity.

The indeterminate form \(0\cdot\infty\)

In a product of a function tending to zero and a function tending to infinity in absolute value, the two factors act in opposite directions: the first tends to make the product small, while the second tends to increase its absolute value.

For example, as \(x\to+\infty\),

\[ \frac{1}{x}\cdot x=1, \qquad \frac{1}{x^2}\cdot x=\frac{1}{x}, \qquad \frac{1}{x}\cdot x^2=x. \]

In all three cases the first factor tends to zero and the second to \(+\infty\), yet the products tend to \(1\), to \(0\), and to \(+\infty\), respectively.

The form \(0\cdot\infty\) is often recast as a quotient. Since a function tending to infinity in absolute value is eventually non-zero, we may write

\[ f(x)g(x) = \frac{f(x)}{\displaystyle\frac{1}{g(x)}}. \]

If \(f(x)\) is also eventually non-zero, one may instead write

\[ f(x)g(x) = \frac{g(x)}{\displaystyle\frac{1}{f(x)}}. \]

The choice of transformation depends on the structure of the expression.

The indeterminate form \(\infty-\infty\)

When two functions tend to the same infinity, their difference depends on how they compare with one another.

For \(x\to+\infty\), consider

\[ (x+1)-x, \qquad x-x, \qquad x-2x. \]

In each case the two terms tend to \(+\infty\), but the differences are, respectively,

\[ 1, \qquad 0, \qquad -x. \]

The limits are therefore \(1\), \(0\), and \(-\infty\).

To remove this form one usually tries to factor out a common term, combine over a common denominator, rationalise, or otherwise rewrite the expression in an equivalent form better suited to the comparison.

Exponential indeterminate forms

When both the base and the exponent depend on \(x\), the forms

\[ 1^\infty, \qquad 0^0, \qquad \infty^0 \]

may also arise.

The form \(1^\infty\) occurs when the base tends to \(1\) while the exponent tends to \(+\infty\) or \(-\infty\). The fact that the base approaches \(1\) is not enough to determine the limit of the power.

For example, as \(x\to+\infty\),

\[ \left(1+\frac{1}{x^2}\right)^x\to 1, \qquad \left(1+\frac{1}{x}\right)^x\to e, \qquad \left(1+\frac{1}{\sqrt{x}}\right)^x\to+\infty. \]

In all three cases the base tends to \(1\) and the exponent to \(+\infty\), yet the limits differ.

The form \(0^0\) can likewise produce different behaviours. For \(x\to 0^+\), consider

\[ x^x\to 1, \qquad x^{\frac{1}{|\ln(x)|}}\to e^{-1}, \qquad x^{\frac{1}{\sqrt{|\ln(x)|}}}\to 0. \]

In all three cases the base tends to \(0^+\) and the exponent to zero. The limits are nonetheless different, since the behaviour of the power depends on the product of the exponent with the natural logarithm of the base.

For the form \(\infty^0\), with \(x\to+\infty\), one may consider

\[ x^{1/x}\to 1, \qquad x^{1/\ln(x)}\to e, \qquad x^{1/\sqrt{\ln(x)}}\to+\infty. \]

In all three cases the base tends to \(+\infty\) and the exponent to zero, yet the limits differ.

Exponential forms are ordinarily studied by taking the natural logarithm. If

\[ y(x)=\bigl(f(x)\bigr)^{g(x)} \]

and \(f(x)>0\) at the points considered, then

\[ \ln\bigl(y(x)\bigr)=g(x)\ln\bigl(f(x)\bigr). \]

The problem is thus reduced to the study of the product \(g(x)\ln\bigl(f(x)\bigr)\).

Summary

The principal indeterminate forms are

\[ \frac{0}{0}, \qquad \frac{\infty}{\infty}, \qquad 0\cdot\infty, \qquad \infty-\infty, \qquad 1^\infty, \qquad 0^0, \qquad \infty^0. \]

Recognising an indeterminate form does not complete the evaluation of the limit: it merely marks the point at which the immediate rules no longer suffice. From then on, one must transform the expression or compare the functions involved more carefully.

Limits of polynomials and rational functions

The theorems on operations allow us to compute directly the limits of polynomials and, when the denominator does not tend to zero, those of rational functions.

The limit of a polynomial at a finite point

Let

\[ P(x)=a_0+a_1x+\cdots+a_nx^n \]

be a polynomial with real coefficients. Since

\[ \lim_{x\to x_0}x=x_0, \]

the theorems on sums, on multiplication by a constant, and on powers give

\[ \lim_{x\to x_0}P(x) = a_0+a_1x_0+\cdots+a_nx_0^n = P(x_0). \]

The limit of a polynomial at a real point is therefore computed by substituting the value \(x_0\) directly for \(x\).

This expresses the continuity of polynomials throughout \(\mathbb{R}\).

The limit of a rational function at a finite point

Let \(P\) and \(Q\) be two polynomials, with \(Q\) not identically zero, and consider the rational function

\[ R(x)=\frac{P(x)}{Q(x)}. \]

If

\[ Q(x_0)\neq 0, \]

then the quotient theorem gives

\[ \lim_{x\to x_0}\frac{P(x)}{Q(x)} = \frac{P(x_0)}{Q(x_0)}. \]

Here too the limit is obtained by direct substitution.

If instead \(Q(x_0)=0\), the quotient theorem cannot be applied. The behaviour depends on the numerator and on the algebraic structure of the fraction.

If

\[ P(x_0)\neq 0 \qquad\text{and}\qquad Q(x)\to 0, \]

the absolute value of the quotient grows without bound, but to determine its sign one must examine the signs of the numerator and the denominator near \(x_0\).

If, on the other hand,

\[ P(x_0)=Q(x_0)=0, \]

one meets the indeterminate form \(\displaystyle\frac{0}{0}\). In this case it may be possible to factorise the polynomials and cancel the common factors.

If \(P\) is the zero polynomial, the rational function equals \(0\) at every point of its domain, and the limit equals \(0\). Suppose, then, that neither \(P\) nor \(Q\) is the zero polynomial, and let \(r,s\geq 1\) denote the respective multiplicities of \(x_0\) as a root. We may then write

\[ P(x)=(x-x_0)^rP_1(x), \qquad Q(x)=(x-x_0)^sQ_1(x), \]

with

\[ P_1(x_0)\neq 0, \qquad Q_1(x_0)\neq 0. \]

For \(x\neq x_0\),

\[ \frac{P(x)}{Q(x)} = (x-x_0)^{r-s}\frac{P_1(x)}{Q_1(x)}. \]

If \(r>s\), then

\[ \lim_{x\to x_0}\frac{P(x)}{Q(x)}=0. \]

If \(r=s\), then

\[ \lim_{x\to x_0}\frac{P(x)}{Q(x)} = \frac{P_1(x_0)}{Q_1(x_0)}. \]

If \(r<s\), finally,

\[ \left|\frac{P(x)}{Q(x)}\right|\to+\infty. \]

In this last case, to decide whether the limit is \(+\infty\), \(-\infty\), or whether the right-hand and left-hand limits differ, one must analyse the sign of the expression.

The limit of a polynomial at infinity

Consider a polynomial of degree \(n\geq 1\):

\[ P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0, \qquad a_n\neq 0. \]

Factoring out the highest-degree power, we obtain

\[ P(x) = x^n \left( a_n+\frac{a_{n-1}}{x} +\cdots+ \frac{a_1}{x^{n-1}} +\frac{a_0}{x^n} \right). \]

As \(x\to+\infty\) or \(x\to-\infty\), every term containing a positive power of \(x\) in the denominator tends to zero. Consequently,

\[ a_n+\frac{a_{n-1}}{x} +\cdots+ \frac{a_0}{x^n} \to a_n. \]

The behaviour of the polynomial is therefore governed by its highest-degree term:

\[ P(x)\sim a_nx^n \qquad \text{as }x\to\pm\infty. \]

The symbol \(\sim\) indicates that the ratio of the two expressions tends to \(1\):

\[ \lim_{x\to\pm\infty}\frac{P(x)}{a_nx^n}=1. \]

As \(x\to+\infty\), the sign of the limit coincides with that of the leading coefficient \(a_n\):

\[ \lim_{x\to+\infty}P(x) = \begin{cases} +\infty, & a_n>0,\\ -\infty, & a_n<0. \end{cases} \]

As \(x\to-\infty\), the parity of the degree also comes into play:

\[ \lim_{x\to-\infty}P(x) = \begin{cases} +\infty, & n\text{ even and }a_n>0,\\ -\infty, & n\text{ even and }a_n<0,\\ -\infty, & n\text{ odd and }a_n>0,\\ +\infty, & n\text{ odd and }a_n<0. \end{cases} \]

The limit of a rational function at infinity

Let

\[ P(x)=a_nx^n+\cdots+a_0, \qquad Q(x)=b_mx^m+\cdots+b_0, \]

with \(a_n\neq 0\) and \(b_m\neq 0\). To study

\[ \lim_{x\to\pm\infty}\frac{P(x)}{Q(x)}, \]

we factor out \(x^n\) in the numerator and \(x^m\) in the denominator:

\[ \begin{aligned} \frac{P(x)}{Q(x)} &= \frac{ x^n\left( a_n+\displaystyle\frac{a_{n-1}}{x}+\cdots+ \displaystyle\frac{a_0}{x^n} \right) }{ x^m\left( b_m+\displaystyle\frac{b_{m-1}}{x}+\cdots+ \displaystyle\frac{b_0}{x^m} \right) }\\ &= x^{n-m} \frac{ a_n+\displaystyle\frac{a_{n-1}}{x}+\cdots+ \displaystyle\frac{a_0}{x^n} }{ b_m+\displaystyle\frac{b_{m-1}}{x}+\cdots+ \displaystyle\frac{b_0}{x^m} }. \end{aligned} \]

The second factor tends to

\[ \frac{a_n}{b_m}. \]

The result therefore depends on the comparison of the degrees \(n\) and \(m\).

If \(n<m\), then \(x^{n-m}\to 0\) and

\[ \lim_{x\to\pm\infty}\frac{P(x)}{Q(x)}=0. \]

If \(n=m\), then

\[ \lim_{x\to\pm\infty}\frac{P(x)}{Q(x)} = \frac{a_n}{b_m}. \]

If \(n>m\), the absolute value of the quotient tends to \(+\infty\):

\[ \left|\frac{P(x)}{Q(x)}\right|\to+\infty. \]

The sign of the limit is determined by examining

\[ \frac{a_n}{b_m}x^{n-m}. \]

In particular, as \(x\to+\infty\) the sign coincides with that of the ratio \(\displaystyle\frac{a_n}{b_m}\). As \(x\to-\infty\), one must also take into account the parity of the difference \(n-m\).

The leading-term principle

The preceding results may be summarised by observing that, at infinity, the lower-degree terms become negligible compared with the highest-degree ones:

\[ P(x)\sim a_nx^n, \qquad Q(x)\sim b_mx^m. \]

Consequently,

\[ \frac{P(x)}{Q(x)} \sim \frac{a_n}{b_m}x^{n-m}. \]

This relation does not replace the proof, but it captures its content: the asymptotic behaviour of a rational function is governed by the highest-degree terms of the numerator and the denominator.

Summary examples and common pitfalls

In evaluating a limit, the first step is to identify the limits of the individual parts of the expression. Only after this check can one decide whether the theorems on operations lead directly to the result or whether an indeterminate form arises.

Sum, product, and powers

Consider

\[ \lim_{x\to 2}\left(3x^2-5x+4\right). \]

Since the function is a polynomial, the limit is obtained by direct substitution:

\[ \begin{aligned} \lim_{x\to 2}\left(3x^2-5x+4\right) &=3\cdot 2^2-5\cdot 2+4\\ &=12-10+4\\ &=6. \end{aligned} \]

The substitution is legitimate because the result follows from the theorems on sums, products, and powers.

Quotient with non-zero denominator

Consider

\[ \lim_{x\to 1}\frac{x^2+2x}{x+3}. \]

The numerator tends to \(3\), while the denominator tends to \(4\neq 0\). We may therefore apply the quotient theorem:

\[ \lim_{x\to 1}\frac{x^2+2x}{x+3} = \frac{3}{4}. \]

The indeterminate form \(\displaystyle\frac{0}{0}\)

Consider

\[ \lim_{x\to 2}\frac{x^2-4}{x-2}. \]

Direct substitution produces the indeterminate form

\[ \frac{0}{0}. \]

We factorise the numerator:

\[ x^2-4=(x-2)(x+2). \]

For \(x\neq 2\), we may cancel the common factor:

\[ \frac{x^2-4}{x-2}=x+2. \]

Since the limit depends on the values taken near \(2\), and not on the value at the point itself, it follows that

\[ \lim_{x\to 2}\frac{x^2-4}{x-2} = \lim_{x\to 2}(x+2) = 4. \]

The indeterminate form \(\infty-\infty\)

Consider

\[ \lim_{x\to+\infty}\left(\sqrt{x^2+x}-x\right). \]

Both terms tend to \(+\infty\), so the indeterminate form \(\infty-\infty\) arises. We multiply and divide by the conjugate expression:

\[ \begin{aligned} \sqrt{x^2+x}-x &= \frac{ \left(\sqrt{x^2+x}-x\right) \left(\sqrt{x^2+x}+x\right) }{ \sqrt{x^2+x}+x }\\ &= \frac{x}{\sqrt{x^2+x}+x}. \end{aligned} \]

Since \(x>0\) for all sufficiently large \(x\), we may divide numerator and denominator by \(x\):

\[ \frac{x}{\sqrt{x^2+x}+x} = \frac{1}{\sqrt{1+\displaystyle\frac{1}{x}}+1}. \]

At this stage the limit is determinate:

\[ \lim_{x\to+\infty} \frac{1}{\sqrt{1+\displaystyle\frac{1}{x}}+1} = \frac{1}{2}. \]

Quotient of polynomials at infinity

Consider

\[ \lim_{x\to-\infty} \frac{2x^3-x+1}{x^2+4}. \]

The degree of the numerator exceeds that of the denominator. The behaviour of the quotient is therefore governed by the ratio of the highest-degree terms:

\[ \frac{2x^3}{x^2}=2x. \]

Since \(2x\to-\infty\) as \(x\to-\infty\), we obtain

\[ \lim_{x\to-\infty} \frac{2x^3-x+1}{x^2+4} = -\infty. \]

The same result may be verified by dividing numerator and denominator by \(x^2\):

\[ \frac{2x^3-x+1}{x^2+4} = \frac{ 2x-\displaystyle\frac{1}{x} +\displaystyle\frac{1}{x^2} }{ 1+\displaystyle\frac{4}{x^2} }. \]

The denominator tends to \(1\), while the numerator tends to \(-\infty\).

Denominator tending to zero

Consider

\[ \lim_{x\to 1}\frac{1}{x-1}. \]

The denominator tends to zero and takes opposite signs to the right and to the left of \(x=1\). For \(x\to 1^+\),

\[ x-1\to 0^+, \]

and hence

\[ \lim_{x\to 1^+}\frac{1}{x-1}=+\infty. \]

For \(x\to 1^-\), on the other hand,

\[ x-1\to 0^-, \]

whence

\[ \lim_{x\to 1^-}\frac{1}{x-1}=-\infty. \]

The right-hand and left-hand limits differ; hence the limit as \(x\to 1\) does not exist, not even as an infinite limit.

Errors to avoid

The first error is to apply the quotient theorem when the limit of the denominator is zero. From

\[ f(x)\to L \qquad\text{and}\qquad g(x)\to 0 \]

no general formula for the limit of \(\displaystyle\frac{f(x)}{g(x)}\) follows. One must distinguish the case \(L\neq 0\) from the case \(L=0\) and, where necessary, examine the sign of the denominator.

A second error is to treat indeterminate forms as results. To write

\[ \frac{0}{0}, \qquad \infty-\infty \qquad\text{or}\qquad 0\cdot\infty \]

does not complete the computation: it merely signals that the expression must be transformed or examined more carefully.

Nor should \(+\infty\) and \(-\infty\) be treated as real numbers. Formulae such as

\[ +\infty+M=+\infty \]

are abbreviations of results on limits, and are meaningful only within the context of the hypotheses that justify them.

Another common error is to neglect the domain. Before computing the limit of an expression containing a root, a denominator, a logarithm, or a real power, one must check that the expression is defined at points of its domain arbitrarily close to the point towards which the variable tends.

Finally, direct substitution is not a universal rule. It is valid when it stems from the continuity of the functions involved and when all the operations remain defined under passage to the limit. If it produces an indeterminate form, it does not give the value sought.

Operations on limits thus preserve, under precise hypotheses, the algebraic structure of functions. Their correct application, however, always requires distinguishing between determinate cases, indeterminate forms, and situations in which the limit does not exist.