Indeterminate Forms in Limits: Theory and Classification

What indeterminate forms are

In computing a limit, one often studies an expression built from several functions. When the limits of the individual functions are known, the rules of the algebra of limits allow us, in many cases, to determine directly the limit of the overall expression. In certain situations, however, the limits of the individual terms lead to a symbolic configuration that does not contain enough information to establish the result. Such a configuration is called an indeterminate form.

For example, let \(f\) and \(g\) be two functions such that

\[ \lim_{x\to x_0}f(x)=0 \qquad\text{and}\qquad \lim_{x\to x_0}g(x)=0, \]

and suppose that the quotient \(\displaystyle\frac{f(x)}{g(x)}\) is defined at points of the domain sufficiently close to \(x_0\). The limit

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

is then said to present the indeterminate form

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

This notation does not mean that the quotient is equal to \(\displaystyle\frac{0}{0}\). Indeed, the symbol \(\displaystyle\frac{0}{0}\) does not represent a number and cannot be treated as an ordinary fraction: it merely indicates that both the numerator and the denominator tend to zero.

The term indeterminate expresses precisely the fact that the separate behaviour of the terms does not determine that of the overall expression. Starting from the same indeterminate form, the limit may take a real value, be infinite, or fail to exist.

An indeterminate form is therefore not the outcome of a limit, but a signal that further analysis of the expression and of the relative behaviour of its terms is required.

Why an indeterminate form does not determine the limit

To understand why an indeterminate form does not allow us to establish the value of a limit, let us consider a few quotients that, as \(x\to 0\), all present the form

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

The first example is

\[ \lim_{x\to 0}\frac{x}{x}. \]

For every \(x\neq 0\) we have

\[ \frac{x}{x}=1, \]

and hence

\[ \lim_{x\to 0}\frac{x}{x}=1. \]

Let us now consider the limit

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

For every \(x\neq 0\) we may cancel the factor \(x\) and obtain

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

Hence

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

The same indeterminate form can also lead to an infinite limit. Indeed,

\[ \lim_{x\to 0}\frac{x^2}{x^4} = \lim_{x\to 0}\frac{1}{x^2} = +\infty. \]

Finally, consider

\[ \lim_{x\to 0}\frac{x}{|x|}. \]

For \(x>0\) we have \(|x|=x\), while for \(x<0\) we have \(|x|=-x\). Consequently,

\[ \frac{x}{|x|} = \begin{cases} 1, & x>0,\\ -1, & x<0. \end{cases} \]

It follows that

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

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

We have thus obtained, starting from the same indeterminate form \(\displaystyle\frac{0}{0}\), four different outcomes:

\[ 1,\qquad 0,\qquad +\infty,\qquad \text{the limit does not exist}. \]

The form \(\displaystyle\frac{0}{0}\) therefore does not determine the limit of the quotient. It merely describes the fact that numerator and denominator both tend to zero, without indicating which of the two approaches zero more rapidly, or what their relative behaviour is.

The same principle holds for all the other indeterminate forms: to establish the limit, one must study the structure of the expression and compare the behaviour of its constituent terms more precisely.

The seven indeterminate forms

The indeterminate forms encountered in the calculation of limits are traditionally seven:

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

These symbols do not represent operations between numbers, but describe the limiting behaviour of the terms making up an expression.

The first two are quotient forms:

• \(\displaystyle\frac{0}{0}\), when the numerator and denominator both tend to zero;
• \(\displaystyle\frac{\infty}{\infty}\), when the numerator and denominator each tend to \(+\infty\) or to \(-\infty\), not necessarily with the same sign.

The third is a product form:

• \(0\cdot\infty\), when one factor tends to zero and the other tends to \(+\infty\) or to \(-\infty\).

The fourth is a difference form:

• \(\infty-\infty\), when the two terms of the difference both tend to \(+\infty\) or both tend to \(-\infty\).

The last three are exponential forms:

• \(1^\infty\), when the base tends to \(1\) and the exponent tends to \(+\infty\) or to \(-\infty\);
• \(0^0\), when the base tends to \(0^+\) and the exponent tends to zero;
• \(\infty^0\), when the base tends to \(+\infty\) and the exponent tends to zero.

In the exponential forms, when working within the set of real numbers, the base is required to be positive at points sufficiently close to the point towards which the variable tends. This condition allows the power to be interpreted by means of the logarithm and its limit to be studied correctly.

Not every combination involving zero or infinity is indeterminate. For instance, under suitable assumptions on the signs,

\[ \frac{a}{\infty}=0, \qquad \frac{\infty}{a}=\pm\infty, \qquad 0+\infty=\infty, \]

where \(a\) denotes a finite real number different from zero. These expressions, too, have a purely symbolic meaning: they summarise properties of limits and must not be interpreted as arithmetic operations performed with infinity.

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

Let \(f\) and \(g\) be two functions such that

\[ \lim_{x\to x_0}f(x)=0 \qquad\text{and}\qquad \lim_{x\to x_0}g(x)=0, \]

and suppose that the quotient \(\displaystyle\frac{f(x)}{g(x)}\) is defined at points of the domain sufficiently close to \(x_0\), possibly excluding \(x_0\) itself. In this case, the limit

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

presents the indeterminate form

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

The fact that the numerator and denominator both tend to zero does not allow us to determine the limit of their ratio. The result depends on the relative behaviour of the two functions, that is, on how quickly they approach zero and on the possible presence of oscillation or changes of sign.

For example, consider

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

The numerator and denominator both tend to zero, so the limit presents the form \(\displaystyle\frac{0}{0}\). However, for \(x\neq 1\), we can factorise the numerator and simplify:

\[ \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1. \]

The functions \(\displaystyle\frac{x^2-1}{x-1}\) and \(x+1\) coincide at every point of the domain sufficiently close to \(1\), possibly excluding the point \(1\) itself. Consequently, they have the same limit as \(x\to 1\), and hence

\[ \lim_{x\to 1}\frac{x^2-1}{x-1} = \lim_{x\to 1}(x+1) = 2. \]

The form \(\displaystyle\frac{0}{0}\) can also arise when it is not possible to cancel a common factor immediately. Consider, for example,

\[ \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x}. \]

Here too the numerator and denominator tend to zero. Multiplying numerator and denominator by the conjugate \(\sqrt{1+x}+1\), for \(x\neq 0\) we obtain

\begin{align} \frac{\sqrt{1+x}-1}{x} &= \frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)} {x(\sqrt{1+x}+1)} \\ &= \frac{x}{x(\sqrt{1+x}+1)} = \frac{1}{\sqrt{1+x}+1}. \end{align}

Hence

\[ \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x} = \lim_{x\to 0}\frac{1}{\sqrt{1+x}+1} = \frac{1}{2}. \]

These examples show that the form \(\displaystyle\frac{0}{0}\) must not be interpreted as a result, but as an indication that the quotient needs to be transformed or analysed more carefully. Depending on the structure of the functions involved, factorisation, simplification, rationalisation, standard limits, asymptotic equivalences, or other tools of the calculus of limits may prove useful.

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

Let \(f\) and \(g\) be two functions and let \(L_f,L_g\in\{-\infty,+\infty\}\) be such that

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

where \(x\to x_0\) may be replaced by \(x\to+\infty\) or by \(x\to-\infty\). Suppose further that the quotient \(\displaystyle\frac{f(x)}{g(x)}\) is defined at points of the domain sufficiently close to the limiting situation under consideration. The limit of the quotient then presents the indeterminate form

\[ \frac{\infty}{\infty}. \]

This notation indicates that the numerator and denominator each tend to one of the two infinities, not necessarily with the same sign. It does not, however, allow us to establish the behaviour of their ratio: the result depends on how rapidly the two functions grow in absolute value, and on their signs.

Consider, for example, the limit

\[ \lim_{x\to+\infty}\frac{3x^2-x+1}{2x^2+5}. \]

Both the numerator and the denominator tend to \(+\infty\), so the limit presents the form \(\displaystyle\frac{\infty}{\infty}\). To compare their behaviour, we divide numerator and denominator by \(x^2\):

\[ \frac{3x^2-x+1}{2x^2+5} = \frac{3-\displaystyle\frac{1}{x}+\displaystyle\frac{1}{x^2}} {2+\displaystyle\frac{5}{x^2}}. \]

Since

\[ \lim_{x\to+\infty}\frac{1}{x}=0 \qquad\text{and}\qquad \lim_{x\to+\infty}\frac{1}{x^2}=0, \]

we obtain

\[ \lim_{x\to+\infty}\frac{3x^2-x+1}{2x^2+5} = \frac{3}{2}. \]

The same indeterminate form can also lead to different results. For instance,

\[ \lim_{x\to+\infty}\frac{x}{x^2} = \lim_{x\to+\infty}\frac{1}{x} = 0, \]

while

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

Moreover,

\[ \lim_{x\to+\infty}\frac{x(2+\sin x)}{x} = \lim_{x\to+\infty}(2+\sin x) \]

does not exist, because the function \(2+\sin x\) continues to oscillate between \(1\) and \(3\) without tending to a single value.

The form \(\displaystyle\frac{\infty}{\infty}\) can therefore lead to a finite real number, to zero, to an infinite limit, or to a non-existent limit. Here too, the indeterminate form does not supply the result: it signals the need to compare more precisely the order of growth and the relative behaviour of the numerator and denominator.

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

Let \(f\) and \(g\) be two functions and let \(L\in\{-\infty,+\infty\}\) be such that

\[ \lim_{x\to x_0}f(x)=0 \qquad\text{and}\qquad \lim_{x\to x_0}g(x)=L, \]

where \(x\to x_0\) may be replaced by another limiting situation, and suppose that the product \(f(x)g(x)\) is defined at points of the domain sufficiently close to the situation under consideration. The limit

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

then presents the indeterminate form

\[ 0\cdot\infty. \]

This notation does not represent the product of the number \(0\) and infinity. It merely indicates that one factor tends to zero, while the other grows without bound in absolute value.

The two factors exert opposing effects: the factor tending to zero tends to make the product small, while the divergent one tends to make it large in absolute value. The limit therefore depends on their relative behaviour.

Consider, for example,

\[ \lim_{x\to+\infty}\frac{1}{x}\cdot x. \]

The first factor tends to zero and the second tends to \(+\infty\), so the limit presents the form \(0\cdot\infty\). However, for every \(x>0\),

\[ \frac{1}{x}\cdot x=1, \]

and hence

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

The same form can lead to zero. Indeed,

\[ \lim_{x\to+\infty}\frac{1}{x^2}\cdot x = \lim_{x\to+\infty}\frac{1}{x} = 0. \]

It can also lead to an infinite limit:

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

Finally, the limit may fail to exist. Consider

\[ \lim_{x\to+\infty}\frac{\sin x}{x}\cdot x. \]

The first factor tends to zero, because

\[ \left|\frac{\sin x}{x}\right|\leq\frac{1}{x} \]

for \(x>0\), while the second tends to \(+\infty\). The product therefore presents the form \(0\cdot\infty\). However, for every \(x>0\),

\[ \frac{\sin x}{x}\cdot x=\sin x, \]

and the limit does not exist, because \(\sin x\) oscillates without tending to a single value.

To study a form \(0\cdot\infty\), the product is generally transformed into a quotient reducible to the form \(\displaystyle\frac{0}{0}\) or, when conditions allow, to the form \(\displaystyle\frac{\infty}{\infty}\). The choice of the most convenient transformation depends on the structure of the functions involved.

The form \(0\cdot\infty\) therefore does not determine the limit, but signals that the product must be rewritten or analysed more carefully.

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

Let \(f\) and \(g\) be two functions such that

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

or

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

where \(x\to x_0\) may be replaced by another limiting situation. The limit of the difference

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

presents the indeterminate form

\[ \infty-\infty. \]

This notation does not represent a subtraction between infinite numbers. It indicates that the two terms of the difference diverge with the same sign, and that, consequently, the behaviour of their difference cannot be established from knowledge of their separate limits alone.

The result depends on the manner in which the two terms diverge and, in particular, on the difference between their orders of growth.

For example,

\[ \lim_{x\to+\infty}\bigl((x+1)-x\bigr)=1. \]

Both terms \(x+1\) and \(x\) tend to \(+\infty\), but their difference is constantly equal to \(1\).

The same form can lead to zero:

\[ \lim_{x\to+\infty}(x-x)=0. \]

It can lead to \(+\infty\):

\[ \lim_{x\to+\infty}(2x-x) = \lim_{x\to+\infty}x = +\infty, \]

or to \(-\infty\):

\[ \lim_{x\to+\infty}(x-2x) = \lim_{x\to+\infty}(-x) = -\infty. \]

The limit may also fail to exist. Indeed,

\[ \lim_{x\to+\infty}\bigl((x+\sin x)-x\bigr) = \lim_{x\to+\infty}\sin x \]

does not exist, even though both \(x+\sin x\to+\infty\) and \(x\to+\infty\).

To resolve a form \(\infty-\infty\), one generally seeks to transform the difference into a single expression, so as to make the relative behaviour of the two terms explicit. Depending on the structure of the expression, it may be useful to factor out a common term, reduce the terms to a common denominator, or rationalise.

Consider, for example,

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

Since

\[ \sqrt{x^2+x}\to+\infty \qquad\text{and}\qquad x\to+\infty, \]

the limit presents the form \(\infty-\infty\). We multiply and divide by the conjugate expression:

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

In the numerator we obtain a difference of squares:

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

Since \(x>0\) eventually as \(x\to+\infty\), we may divide numerator and denominator by \(x\):

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

Hence

\[ \lim_{x\to+\infty}\left(\sqrt{x^2+x}-x\right) = \frac{1}{2}. \]

The form \(\infty-\infty\) therefore does not determine the limit of the difference. To obtain the result it is necessary to transform the expression and bring out the actual comparison between the diverging terms.

The indeterminate form \(1^\infty\)

Let \(f\) and \(g\) be two functions and let \(L\in\{-\infty,+\infty\}\) be such that

\[ \lim_{x\to x_0}f(x)=1 \qquad\text{and}\qquad \lim_{x\to x_0}g(x)=L, \]

where \(x\to x_0\) may be replaced by another limiting situation. Since \(f(x)\to 1\), we have \(f(x)>0\) eventually; the power \(f(x)^{g(x)}\) can therefore be studied, within the reals, by means of the logarithm.

The limit

\[ \lim_{x\to x_0}f(x)^{g(x)} \]

then presents the indeterminate form

\[ 1^\infty. \]

This notation does not represent a power with base \(1\) and infinite exponent. It merely indicates that the base tends to \(1\), while the exponent tends to \(+\infty\) or to \(-\infty\).

The form is indeterminate because two effects are in competition. On the one hand, a base close to \(1\) tends to alter the value of the power only slightly; on the other, an exponent very large in absolute value can make even a very small difference between the base and \(1\) decisive. The result therefore depends on the joint behaviour of the base and the exponent.

To analyse this form, set

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

Since \(f(x)>0\), we may apply the natural logarithm:

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

Since \(f(x)\to 1\), by continuity of the logarithm we have

\[ \ln f(x)\to\ln 1=0. \]

The product

\[ g(x)\ln f(x) \]

therefore presents the form \(0\cdot\infty\). The study of the exponential form \(1^\infty\) is thus reduced to the study of the limit of the logarithm of the expression.

In particular, if

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

with \(L\in\mathbb{R}\), then, by continuity of the exponential function,

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

If instead

\[ g(x)\ln f(x)\to+\infty, \]

then

\[ f(x)^{g(x)}\to+\infty; \]

if

\[ g(x)\ln f(x)\to-\infty, \]

then

\[ f(x)^{g(x)}\to 0. \]

Consider the standard limit

\[ \lim_{x\to+\infty}\left(1+\frac{1}{x}\right)^x=e. \]

The base tends to \(1\) and the exponent tends to \(+\infty\), yet the limit is the positive real number \(e\).

The same form can lead to \(1\). Indeed,

\[ \left(1+\frac{1}{x}\right)^{\sqrt{x}} = \left[\left(1+\frac{1}{x}\right)^x\right]^{\frac{1}{\sqrt{x}}} \longrightarrow e^0=1 \qquad\text{as }x\to+\infty. \]

It can also lead to \(+\infty\):

\[ \left(1+\frac{1}{x}\right)^{x^2} = \left[\left(1+\frac{1}{x}\right)^x\right]^x \longrightarrow+\infty. \]

A result equal to zero is obtained, for example, in the limit

\[ \lim_{x\to+\infty}\left(1-\frac{1}{x}\right)^{x^2}. \]

For \(x>1\), the base is positive and we may write

\[ \left(1-\frac{1}{x}\right)^{x^2} = \left[\left(1-\frac{1}{x}\right)^x\right]^x. \]

Since

\[ \lim_{x\to+\infty}\left(1-\frac{1}{x}\right)^x=e^{-1}, \]

and \(0<e^{-1}<1\), it follows that

\[ \lim_{x\to+\infty}\left(1-\frac{1}{x}\right)^{x^2}=0. \]

Finally, the limit may fail to exist. Consider

\[ \lim_{x\to+\infty}\left(1+\frac{1}{x}\right)^{x(2+\sin x)}. \]

The base tends to \(1\), while the exponent \(x(2+\sin x)\) tends to \(+\infty\), since \(2+\sin x\geq 1\). Taking the logarithm, we obtain

\[ \ln\left[\left(1+\frac{1}{x}\right)^{x(2+\sin x)}\right] = (2+\sin x)\,x\ln\left(1+\frac{1}{x}\right). \]

Since

\[ x\ln\left(1+\frac{1}{x}\right)\to 1, \]

along the sequence \(x_n=\displaystyle\frac{\pi}{2}+2\pi n\) the logarithm of the expression tends to \(3\), while along the sequence \(y_n=\displaystyle\frac{3\pi}{2}+2\pi n\) it tends to \(1\). Consequently, the expression tends to \(e^3\) and to \(e\) respectively, and hence the limit does not exist.

The form \(1^\infty\) can therefore lead to \(1\), to a positive real number different from \(1\), to zero, to \(+\infty\), or to a non-existent limit. Its study requires considering the product of the exponent and the logarithm of the base.

The indeterminate form \(0^0\)

Let \(f\) and \(g\) be two functions such that

\[ \lim_{x\to x_0}f(x)=0 \qquad\text{and}\qquad \lim_{x\to x_0}g(x)=0, \]

where \(x\to x_0\) may be replaced by another limiting situation. Suppose further that \(f(x)>0\) at points of the domain sufficiently close to the situation under consideration. The limit

\[ \lim_{x\to x_0}f(x)^{g(x)} \]

then presents the indeterminate form

\[ 0^0. \]

The notation \(0^0\), in this context, does not represent the value taken by the power and must not be interpreted as an operation between two numbers. It merely indicates that the base tends to zero, while the exponent tends to zero.

The form is indeterminate because the behaviour of the power depends on the manner in which the base and the exponent approach their respective limiting values. A very small positive base tends to make the power small when the exponent is positive, while an exponent tending to zero tends to bring the power close to \(1\). If the exponent takes negative values, the power may instead become arbitrarily large.

To study this form, set \(y=f(x)^{g(x)}\). Since \(f(x)>0\), we may apply the natural logarithm and obtain

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

Since \(f(x)\to 0^+\), we have \(\ln f(x)\to-\infty\); since at the same time \(g(x)\to 0\), the product \(g(x)\ln f(x)\) presents the indeterminate form \(0\cdot\infty\).

If \(g(x)\ln f(x)\) tends to a value \(L\in\mathbb{R}\), to \(-\infty\), or to \(+\infty\), then \(f(x)^{g(x)}\) tends respectively to \(e^L\), to \(0\), or to \(+\infty\).

Consider first the limit

\[ \lim_{x\to 0^+}x^x. \]

The base and exponent both tend to zero. Taking the logarithm, we obtain

\[ \ln(x^x)=x\ln x. \]

Since

\[ \lim_{x\to 0^+}x\ln x=0, \]

it follows that

\[ \lim_{x\to 0^+}x^x=e^0=1. \]

The same form can lead to a positive real number different from \(1\). For example, for \(0<x<1\),

\[ x^{\frac{1}{\ln x}} = e^{\frac{1}{\ln x}\ln x} = e. \]

Since \(x\to 0^+\) and \(\displaystyle\frac{1}{\ln x}\to 0\), the limit presents the form \(0^0\), yet we have

\[ \lim_{x\to 0^+}x^{\frac{1}{\ln x}}=e. \]

The form \(0^0\) can also lead to zero. Consider

\[ \lim_{x\to 0^+}x^{\frac{1}{\sqrt{-\ln x}}}. \]

The exponent tends to zero, and

\[ \ln\left(x^{\frac{1}{\sqrt{-\ln x}}}\right) = \frac{\ln x}{\sqrt{-\ln x}} = -\sqrt{-\ln x} \to-\infty. \]

Hence

\[ \lim_{x\to 0^+}x^{\frac{1}{\sqrt{-\ln x}}}=0. \]

If we change the sign of the exponent, we instead obtain

\[ \ln\left(x^{-\frac{1}{\sqrt{-\ln x}}}\right) = -\frac{\ln x}{\sqrt{-\ln x}} = \sqrt{-\ln x} \to+\infty, \]

and hence

\[ \lim_{x\to 0^+}x^{-\frac{1}{\sqrt{-\ln x}}}=+\infty. \]

Finally, the limit may fail to exist. Consider

\[ x^{\frac{\sin(-\ln x)}{\ln x}}. \]

As \(x\to 0^+\), the base tends to zero and the exponent tends to zero, since the numerator is bounded while \(\ln x\to-\infty\). However,

\[ x^{\frac{\sin(-\ln x)}{\ln x}} = e^{\sin(-\ln x)}. \]

Since \(-\ln x\to+\infty\), the function \(\sin(-\ln x)\) continues to oscillate and has no limit. Consequently, the expression \(e^{\sin(-\ln x)}\) has no limit either.

The form \(0^0\) can therefore lead to \(1\), to a positive real number different from \(1\), to zero, to \(+\infty\), or to a non-existent limit. The result is determined by the behaviour of the product of the exponent and the logarithm of the base.

In certain areas of mathematics, such as combinatorics and the theory of power series, it may be convenient to adopt the convention \(0^0=1\). This convention pertains to specific contexts and does not alter the fact that, in the calculus of limits, \(0^0\) is an indeterminate form.

The indeterminate form \(\infty^0\)

Let \(f\) and \(g\) be two functions such that

\[ \lim_{x\to x_0}f(x)=+\infty \qquad\text{and}\qquad \lim_{x\to x_0}g(x)=0, \]

where \(x\to x_0\) may be replaced by another limiting situation. Since \(f(x)\to+\infty\), the base is positive at points of the domain sufficiently close to the situation under consideration. The limit

\[ \lim_{x\to x_0}f(x)^{g(x)} \]

then presents the indeterminate form

\[ \infty^0. \]

This notation does not represent a power with infinite base and zero exponent. It merely indicates that the base grows without bound, while the exponent tends to zero.

The form is indeterminate because the two behaviours produce opposing effects. A very large base can make the power large, while an exponent close to zero tends to bring the power close to \(1\). If the exponent takes negative values, the power may instead approach zero.

To study this form, set \(y=f(x)^{g(x)}\). Since \(f(x)>0\) eventually, we may apply the natural logarithm and obtain

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

Since \(f(x)\to+\infty\), we have \(\ln f(x)\to+\infty\); since at the same time \(g(x)\to 0\), the product \(g(x)\ln f(x)\) presents the indeterminate form \(0\cdot\infty\).

If \(g(x)\ln f(x)\) tends to a value \(L\in\mathbb{R}\), to \(-\infty\), or to \(+\infty\), then \(f(x)^{g(x)}\) tends respectively to \(e^L\), to \(0\), or to \(+\infty\).

Consider first the limit

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

The base tends to \(+\infty\), while the exponent tends to zero. Taking the logarithm, we obtain

\[ \ln\left(x^{\frac{1}{x}}\right) = \frac{\ln x}{x}. \]

Since

\[ \lim_{x\to+\infty}\frac{\ln x}{x}=0, \]

it follows that

\[ \lim_{x\to+\infty}x^{\frac{1}{x}}=e^0=1. \]

The same form can lead to a positive real number different from \(1\). For example, for \(x>1\),

\[ x^{\frac{1}{\ln x}} = e^{\frac{1}{\ln x}\ln x} = e. \]

Since \(x\to+\infty\) and \(\displaystyle\frac{1}{\ln x}\to 0\), we thus have

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

The form \(\infty^0\) can also lead to zero. Consider

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

The exponent tends to zero, and

\[ \ln\left(x^{-\frac{1}{\sqrt{\ln x}}}\right) = -\frac{\ln x}{\sqrt{\ln x}} = -\sqrt{\ln x} \to-\infty. \]

Hence

\[ \lim_{x\to+\infty}x^{-\frac{1}{\sqrt{\ln x}}}=0. \]

If we change the sign of the exponent, we instead obtain

\[ \ln\left(x^{\frac{1}{\sqrt{\ln x}}}\right) = \sqrt{\ln x} \to+\infty, \]

and hence

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

Finally, the limit may fail to exist. Consider, for \(x>1\),

\[ x^{\frac{\sin(\ln\ln x)}{\ln x}}. \]

As \(x\to+\infty\), the base tends to \(+\infty\) and the exponent tends to zero, since the numerator is bounded while \(\ln x\to+\infty\). However,

\[ x^{\frac{\sin(\ln\ln x)}{\ln x}} = e^{\sin(\ln\ln x)}. \]

Since \(\ln\ln x\to+\infty\), the function \(\sin(\ln\ln x)\) continues to oscillate and has no limit. Consequently, the expression \(e^{\sin(\ln\ln x)}\) has no limit either.

The form \(\infty^0\) can therefore lead to \(1\), to a positive real number different from \(1\), to zero, to \(+\infty\), or to a non-existent limit. Here too, the result is determined by the behaviour of the product of the exponent and the logarithm of the base.

Transforming indeterminate forms

Indeterminate forms are not resolved by arithmetic rules applied directly to the symbols \(0\) and \(\infty\). To determine their limit, it is necessary to rewrite the expression in a form identical to the original for all values of the variable sufficiently close to the limiting situation under consideration.

If \(x\to x_0\), it suffices that the two expressions coincide at points of the domain sufficiently close to \(x_0\), possibly excluding the point \(x_0\) itself. If instead \(x\to+\infty\) or \(x\to-\infty\), it suffices that they coincide, respectively, for all sufficiently large or sufficiently small values of \(x\).

The forms \(\displaystyle\frac{0}{0}\) and \(\displaystyle\frac{\infty}{\infty}\) are already expressed as quotients. The other indeterminate forms can often be reduced to one of these by suitable transformations.

Transforming the form \(0\cdot\infty\)

If

\[ f(x)\to 0 \qquad\text{and}\qquad g(x)\to L, \qquad L\in\{-\infty,+\infty\}, \]

then \(g(x)\neq 0\) eventually and

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

We may therefore write

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

obtaining a form \(\displaystyle\frac{0}{0}\).

If, moreover, \(f(x)\neq 0\) and has constant sign eventually, then \(\displaystyle\frac{1}{f(x)}\) tends to \(+\infty\) or to \(-\infty\), and we may also write

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

obtaining a form \(\displaystyle\frac{\infty}{\infty}\).

The choice between the two rewritings depends on the structure of the functions and on which form proves easier to analyse.

Transforming the form \(\infty-\infty\)

A difference presenting the form \(\infty-\infty\) must be transformed into a single expression, so as to make the relative behaviour of the two diverging terms visible.

If the terms are fractions, it is often useful to reduce them to a common denominator:

\[ \frac{f(x)}{g(x)}-\frac{h(x)}{k(x)} = \frac{f(x)k(x)-h(x)g(x)}{g(x)k(x)}. \]

The form \(\infty-\infty\) can thus be transformed into a quotient form, generally \(\displaystyle\frac{0}{0}\) or \(\displaystyle\frac{\infty}{\infty}\).

When radicals appear, it may be useful to multiply and divide by the conjugate expression. For example,

\[ \sqrt{f(x)}-\sqrt{g(x)} = \frac{f(x)-g(x)} {\sqrt{f(x)}+\sqrt{g(x)}}, \]

at points where the roots are defined and the denominator is different from zero.

In other cases it may be convenient to factor out a common term:

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

provided \(f(x)\neq 0\) at the points under consideration.

Transforming the exponential forms

The forms \(1^\infty\), \(0^0\) and \(\infty^0\) are studied by means of the logarithm. If \(f(x)>0\) eventually, setting

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

we obtain

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

The limit of the power is thus reduced to the study of the product \(g(x)\ln f(x)\). If this tends to a value \(L\in\mathbb{R}\), to \(+\infty\), or to \(-\infty\), then \(f(x)^{g(x)}\) tends respectively to \(e^L\), to \(+\infty\), or to \(0\).

Conversely, if the product \(g(x)\ln f(x)\) admits neither a real limit nor an infinite limit, then the power admits no limit either: otherwise, applying the logarithm to its supposed positive, zero, or infinite limit would yield a limit for \(g(x)\ln f(x)\).

In every transformation it is essential to verify the conditions of validity of the operations performed. An algebraic transformation is useful only when it preserves the expression at points of the domain sufficiently close to the limiting situation and leads to a form that is easier to analyse.

Methods for resolving indeterminate forms

There is no single procedure for determining every limit that presents an indeterminate form. The method to be used depends on the structure of the expression, on the type of indeterminate form, and on the mathematical tools available.

Before performing any transformation, it is necessary to identify the domain of the expression, compute separately the limits of its constituent terms, and verify that an indeterminate form is indeed present.

Algebraic simplification

When a limit presents the form \(\displaystyle\frac{0}{0}\), it may be possible to factorise the numerator and denominator, identify a common factor, and cancel it.

Consider, for example,

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

Using the factorisation of a difference of cubes, for \(x\neq 1\) we obtain

\[ \frac{x^3-1}{x-1} = \frac{(x-1)(x^2+x+1)}{x-1} = x^2+x+1. \]

Hence

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

The simplification is valid because it is performed for \(x\neq 1\), that is, at points sufficiently close to the point towards which the variable tends. The value possibly taken by the expression at the point \(x=1\) does not affect the limit.

Reduction to a common denominator and rationalisation

Differences presenting the form \(\infty-\infty\) can often be transformed by reducing the terms to a common denominator. If radicals appear instead, it may be useful to multiply and divide by the conjugate expression.

These procedures transform the difference into a single fraction, in which it becomes possible to study the overall behaviour of the numerator and denominator.

Here too it is necessary to verify that the denominators introduced are different from zero at the points under consideration and that all roots present are defined.

Dividing by the term of highest degree

In quotients of polynomials, as \(x\to+\infty\) or \(x\to-\infty\), the behaviour is determined by the terms of highest degree. One may therefore divide the numerator and denominator by the highest power of \(x\) present in the denominator or, more generally, by a power that allows the dominant terms to be compared.

For example,

\[ \lim_{x\to+\infty}\frac{5x^3-x}{2x^3+4x} = \lim_{x\to+\infty} \frac{5-\displaystyle\frac{1}{x^2}} {2+\displaystyle\frac{4}{x^2}} = \frac{5}{2}. \]

If radicals are present, particular attention must be paid to the identity

\[ \sqrt{x^2}=|x|. \]

In particular, for \(x\to-\infty\) we have \(|x|=-x\), not \(|x|=x\).

Standard limits

Many indeterminate forms can be reduced to standard limits by algebraic transformations or changes of variable. Among the principal standard limits we have

\[ \lim_{t\to 0}\frac{\sin t}{t}=1, \qquad \lim_{t\to 0}\frac{e^t-1}{t}=1, \]

\[ \lim_{t\to 0}\frac{\ln(1+t)}{t}=1, \qquad \lim_{t\to 0}(1+t)^{\frac{1}{t}}=e. \]

To use a standard limit correctly, the whole expression must be brought into its structure, checking that the inner variable tends to the required value.

For example,

\[ \lim_{x\to 0}\frac{\sin(3x)}{x} = 3\lim_{x\to 0}\frac{\sin(3x)}{3x} = 3. \]

Asymptotic equivalences

Two functions \(f\) and \(g\) are said to be asymptotically equivalent as \(x\to x_0\) if

\[ \lim_{x\to x_0}\frac{f(x)}{g(x)}=1, \]

and one writes

\[ f(x)\sim g(x) \qquad\text{as }x\to x_0. \]

Asymptotic equivalences allow one function to be replaced by another with the same leading behaviour in products and quotients, provided the expressions involved are defined and the denominators are eventually different from zero.

For example, since

\[ \sin x\sim x \qquad\text{as }x\to 0, \]

we have

\[ \frac{\sin x}{x}\sim\frac{x}{x}=1. \]

Equivalences cannot, however, be used indiscriminately in sums and differences, because cancellation of the leading terms may occur.

For example,

\[ x+x^2\sim x \qquad\text{and}\qquad x\sim x \]

as \(x\to 0\), but

\[ (x+x^2)-x=x^2, \]

whereas the separate substitution of the two terms by \(x\) would erroneously give \(x-x=0\). When cancellation occurs, it is necessary to retain terms of higher order.

Comparison theorems

When an expression contains oscillating factors or is not easily simplified, it may be useful to compare it with functions of known behaviour.

Consider, for example,

\[ \lim_{x\to 0}\frac{x^2\sin\left(\displaystyle\frac{1}{x}\right)}{x}. \]

For \(x\neq 0\), the expression reduces to

\[ x\sin\left(\frac{1}{x}\right). \]

Since

\[ \left|x\sin\left(\frac{1}{x}\right)\right| \leq |x|, \]

and \(|x|\to 0\), the squeeze theorem gives

\[ \lim_{x\to 0}x\sin\left(\frac{1}{x}\right)=0. \]

The choice of method must therefore be guided by the structure of the expression. An effective transformation does not formally eliminate the symbols of the indeterminate form, but makes the relative behaviour of the functions producing it explicit.

Common mistakes

In the calculus of indeterminate forms it is essential to distinguish mathematically valid transformations from purely formal manipulations. The most common errors generally arise from an improper interpretation of the symbols \(0\) and \(\infty\), from the automatic application of a rule, or from a failure to verify the necessary conditions.

Regarding the indeterminate form as the result of the limit

Writing that a limit presents the form \(\displaystyle\frac{0}{0}\) does not mean that its value has been determined. The form merely indicates that the numerator and denominator both tend to zero.

For example,

\[ \lim_{x\to 0}\frac{x}{x}=1, \qquad \lim_{x\to 0}\frac{x^2}{x}=0, \qquad \lim_{x\to 0}\frac{x}{x^2} \]

all present the form \(\displaystyle\frac{0}{0}\), yet they produce different behaviours. In the last case, moreover, the two-sided limit does not exist: the right-hand limit is \(+\infty\), while the left-hand limit is \(-\infty\).

Treating infinity as a real number

The symbols \(+\infty\) and \(-\infty\) do not represent real numbers. Expressions such as

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

are not arithmetic operations performed with infinity, but concise descriptions of the limiting behaviour of the terms.

It is therefore not legitimate, for example, formally to cancel the infinities in the expression \(\displaystyle\frac{\infty}{\infty}\) and conclude that the result is \(1\).

Confusing substitution with the computation of the limit

Direct substitution is a useful tool when the operations involved preserve the limit and do not produce indeterminate configurations. If, on the other hand, substitution leads to an indeterminate form, it does not determine the limit, but signals the need for further analysis.

Moreover, the limit of a function at a point does not necessarily depend on the value taken by the function at that point. For this reason, an expression may be undefined at the point towards which the variable tends and nonetheless possess a limit.

Simplifying a factor without specifying the conditions

In the relation

\[ \frac{(x-x_0)h(x)}{x-x_0}=h(x), \]

the simplification is valid only for \(x\neq x_0\). This is sufficient for the computation of the limit as \(x\to x_0\), because the limit depends on the values taken at points of the domain sufficiently close to \(x_0\), possibly excluding \(x_0\) itself.

It would not be correct, however, to state that the two expressions are equal also at \(x=x_0\), since the left-hand side is undefined at that point.

Applying asymptotic equivalences in differences without checking for cancellation

Asymptotic equivalences can be used directly in products and quotients, but cannot be substituted indiscriminately within sums and differences.

For example, as \(x\to 0\), we have

\[ x+x^2\sim x. \]

However, substituting \(x+x^2\) with \(x\) in the difference \((x+x^2)-x\) would give zero, whereas

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

When the leading terms cancel, it is therefore necessary to analyse the expression more precisely and retain the terms that actually determine its behaviour.

Neglecting the domain of the expression

Before computing a limit it is necessary to establish where the expression is defined. This check is particularly important in the presence of denominators, radicals, logarithms, and powers with real exponent.

For example, in the study of

\[ f(x)^{g(x)} \]

by means of the relation

\[ f(x)^{g(x)}=e^{g(x)\ln f(x)}, \]

it is necessary that \(f(x)>0\) at points sufficiently close to the limiting situation under consideration.

Forgetting the absolute value when extracting a square root

The correct identity is

\[ \sqrt{x^2}=|x|, \]

not \(\sqrt{x^2}=x\) for every \(x\in\mathbb{R}\).

Consequently, for \(x\to-\infty\),

\[ \sqrt{x^2}=|x|=-x. \]

Ignoring this fact can change the sign of the expression and lead to an incorrect result.

Confusing an infinite limit with a non-existent limit

To say that

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

expresses a precise behaviour: the values of the function become greater than any prescribed real number when \(x\) is sufficiently close to \(x_0\).

This case differs from one in which the limit fails to exist through oscillation, or because the two one-sided limits differ.

For example,

\[ \lim_{x\to 0}\frac{1}{x^2}=+\infty, \]

whereas

\[ \lim_{x\to 0}\frac{1}{x} \]

does not exist as a two-sided limit, since

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

Neglecting the sign of diverging terms

Knowing only the divergence in absolute value is not always sufficient to determine the sign of the result. In quotients, products, and differences it is also necessary to study the sign eventually taken by the individual terms.

For example, from the sole information

\[ |f(x)|\to+\infty \]

it does not follow that \(f(x)\to+\infty\): the function could tend to \(-\infty\), or oscillate while taking arbitrarily large values in absolute value.

Stopping when a transformation produces another indeterminate form

An indeterminate form can be transformed into another indeterminate form. This does not mean that the transformation is useless or incorrect: it may be a necessary intermediate step.

For example, a form \(0\cdot\infty\) can be rewritten as \(\displaystyle\frac{0}{0}\), while an exponential form can be reduced to \(0\cdot\infty\) by means of the logarithm.

The procedure must continue until the expression obtained allows the properties of limits, or one of the available tools, to be applied correctly.

Avoiding these mistakes requires treating every indeterminate form as a problem of comparison between behaviours, not as a symbolic calculation between \(0\) and \(\infty\). Every step must be justified by the domain, by the properties of the functions involved, and by the hypotheses of the method used.

Summary table

The seven indeterminate forms can be classified as quotient forms, product forms, difference forms, and exponential forms. The table below summarises their meaning and the transformations most frequently used.

For the exponential forms, assuming \(f(x)>0\) eventually, one uses the relation

\[ f(x)^{g(x)}=e^{g(x)\ln f(x)}, \]

thereby reducing the limit of the power to the study of the product \(g(x)\ln f(x)\).

The general procedure for studying a limit can be summarised in the following steps:

1. determine the domain of the expression and specify the limiting situation under consideration;
2. compute separately the limits of the terms making up the expression;
3. verify whether the properties of the algebra of limits allow a direct conclusion, or whether an indeterminate form appears;
4. transform the expression by means of a procedure valid within its domain;
5. apply the most suitable method, verifying all its hypotheses;
6. check the result, paying attention to signs, one-sided limits, and the possible presence of oscillation.

An indeterminate form is therefore not a purely formal obstacle, but a signal that the behaviour of the individual terms is insufficient to determine that of the overall expression. Computing the limit requires bringing out their relative behaviour by means of mathematically justified transformations and tools.