The exercises on this page provide a progressive application of the principal limit laws. We begin with cases in which the result follows directly from the laws for sums, products, quotients, powers and roots, together with the limit theorem for composite functions.
We then turn to situations in which direct substitution yields an indeterminate form. In such cases, merely writing \(\displaystyle\frac{0}{0}\), \(0\cdot\infty\) or \(\infty-\infty\) does not determine the limit: the expression must be transformed by factorisation, cancellation, rationalisation or division by the highest relevant power of the variable.
Particular attention will be paid to domains and signs. When a denominator tends to zero, its limit alone is insufficient: to determine whether the quotient tends to \(+\infty\), to \(-\infty\) or has no limit, one must establish the sign of the denominator to the right and to the left of the point under consideration.
In each exercise, the properties used are stated explicitly and every step is justified. Neither l’Hôpital’s rule nor Taylor expansions are used; every result is obtained solely from the limit laws and algebraic techniques introduced in the theoretical discussion.
Exercise 1 — level ★☆☆☆☆
Evaluate the following limit:
\[ \lim_{x\to 2}\left(3x^2-5x+4\right). \]
Result
\[ \lim_{x\to 2}\left(3x^2-5x+4\right)=6. \]
Solution
The expression \(3x^2-5x+4\) is a polynomial. Since every polynomial is continuous on \(\mathbb{R}\), its limit at a finite point is obtained by directly substituting the value to which the variable tends.
We may also justify the substitution by means of the theorems on operations with limits. Indeed, we know that
\[ \lim_{x\to 2}x=2. \]
By the power rule it follows that
\[ \lim_{x\to 2}x^2=2^2=4. \]
Taking the constant factors outside the corresponding limits, we obtain
\[ \lim_{x\to 2}3x^2=3\cdot 4=12 \]
and
\[ \lim_{x\to 2}(-5x)=-5\cdot 2=-10. \]
The constant function \(4\) has limit \(4\). Hence, by the sum rule,
\[ \begin{aligned} \lim_{x\to 2}\left(3x^2-5x+4\right) &=3\cdot 2^2-5\cdot 2+4\\ &=12-10+4\\ &=6. \end{aligned} \]
Hence the required limit is
\[ 6. \]
Exercise 2 — level ★☆☆☆☆
Let \(f\) and \(g\) be two functions such that
\[ \lim_{x\to x_0}f(x)=-2 \qquad\text{and}\qquad \lim_{x\to x_0}g(x)=3. \]
Evaluate
\[ \lim_{x\to x_0}\left(4f(x)-2g(x)+5\right). \]
Result
\[ \lim_{x\to x_0}\left(4f(x)-2g(x)+5\right)=-9. \]
Solution
Both given limits are finite. We may therefore apply the sum rule and the constant multiple rule.
From the limit of \(f\) it follows that
\[ \lim_{x\to x_0}4f(x) = 4\lim_{x\to x_0}f(x) = 4\cdot(-2) = -8. \]
Similarly,
\[ \lim_{x\to x_0}\bigl(-2g(x)\bigr) = -2\lim_{x\to x_0}g(x) = -2\cdot 3 = -6. \]
The constant function \(5\) has limit equal to \(5\). We may therefore add the three results:
\[ \begin{aligned} \lim_{x\to x_0}\left(4f(x)-2g(x)+5\right) &= 4(-2)-2(3)+5\\ &=-8-6+5\\ &=-9. \end{aligned} \]
The required limit is therefore
\[ -9. \]
Exercise 3 — level ★☆☆☆☆
Evaluate the limit
\[ \lim_{x\to-1}(2x-3)^2(x+4). \]
Result
\[ \lim_{x\to-1}(2x-3)^2(x+4)=75. \]
Solution
The expression is the product of two functions:
\[ (2x-3)^2 \qquad\text{and}\qquad x+4. \]
We compute their limits separately.
Since
\[ \lim_{x\to-1}(2x-3)=2(-1)-3=-5, \]
the power rule gives
\[ \lim_{x\to-1}(2x-3)^2=(-5)^2=25. \]
For the second factor,
\[ \lim_{x\to-1}(x+4)=-1+4=3. \]
Both limits are finite. We may therefore apply the product rule:
\[ \begin{aligned} \lim_{x\to-1}(2x-3)^2(x+4) &= \left(\lim_{x\to-1}(2x-3)^2\right) \left(\lim_{x\to-1}(x+4)\right)\\ &=25\cdot 3\\ &=75. \end{aligned} \]
Hence
\[ 75. \]
Exercise 4 — level ★☆☆☆☆
Evaluate the limit
\[ \lim_{x\to 1}\frac{x^2+2x+3}{2x+5}. \]
Result
\[ \lim_{x\to 1}\frac{x^2+2x+3}{2x+5} = \frac{6}{7}. \]
Solution
The numerator and denominator are polynomials. We may therefore compute their limits separately by direct substitution.
For the numerator,
\[ \lim_{x\to 1}(x^2+2x+3) = 1^2+2\cdot 1+3 = 6. \]
For the denominator,
\[ \lim_{x\to 1}(2x+5) = 2\cdot 1+5 = 7. \]
The limit of the denominator is non-zero:
\[ 7\neq 0. \]
The essential hypothesis of the quotient rule is thus satisfied. We may write
\[ \begin{aligned} \lim_{x\to 1}\frac{x^2+2x+3}{2x+5} &= \frac{\displaystyle\lim_{x\to 1}(x^2+2x+3)} {\displaystyle\lim_{x\to 1}(2x+5)}\\ &= \frac{6}{7}. \end{aligned} \]
The result is
\[ \frac{6}{7}. \]
Exercise 5 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{x\to 2}|x^2-5|. \]
Result
\[ \lim_{x\to 2}|x^2-5|=1. \]
Solution
Let us first consider the function enclosed within the absolute value:
\[ f(x)=x^2-5. \]
Being a polynomial, its limit is computed by direct substitution:
\[ \lim_{x\to 2}(x^2-5)=2^2-5=4-5=-1. \]
The theorem on the limit of an absolute value states that, if \(f(x)\to L\), then
\[ |f(x)|\to |L|. \]
In our case \(L=-1\). Hence,
\[ \lim_{x\to 2}|x^2-5| = |-1| = 1. \]
We may also verify the result by means of the sign-preservation property of limits. Since \(x^2-5\to-1<0\), the function \(x^2-5\) is negative for every \(x\) sufficiently close to \(2\). In such a neighbourhood,
\[ |x^2-5|=-(x^2-5)=5-x^2. \]
Consequently,
\[ \lim_{x\to 2}|x^2-5| = \lim_{x\to 2}(5-x^2) = 5-4 = 1. \]
The required limit is thus
\[ 1. \]
Exercise 6 — level ★★☆☆☆
Evaluate the right-hand limit
\[ \lim_{x\to 0^+}\sqrt{x(1+x)}. \]
Result
\[ \lim_{x\to 0^+}\sqrt{x(1+x)}=0. \]
Solution
Before applying the theorems on limits, we must check the domain of the square root. It is necessary that
\[ x(1+x)\geq 0. \]
For \(x\) positive and sufficiently close to \(0\), both factors \(x\) and \(1+x\) are positive. The expression under the root is therefore nonnegative and the function is defined.
Let us now compute the limit of the radicand. As \(x\to 0^+\),
\[ x\to 0 \qquad\text{and}\qquad 1+x\to 1. \]
By the product rule,
\[ x(1+x)\to 0\cdot 1=0. \]
The square-root function is continuous at \(0\) on its domain. We may therefore pass the limit through the root:
\[ \lim_{x\to 0^+}\sqrt{x(1+x)} = \sqrt{\lim_{x\to 0^+}x(1+x)} = \sqrt{0} = 0. \]
Hence
\[ 0. \]
Exercise 7 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{x\to-3}\sqrt[3]{2x+5}. \]
Result
\[ \lim_{x\to-3}\sqrt[3]{2x+5}=-1. \]
Solution
The cube root has odd index and is therefore defined for every real number. No nonnegativity condition need be imposed on the radicand.
Let us compute the limit of the inner expression:
\[ \lim_{x\to-3}(2x+5) = 2(-3)+5 = -6+5 = -1. \]
The function
\[ \varphi(t)=\sqrt[3]{t} \]
is continuous on the whole of \(\mathbb{R}\). We may therefore apply the theorem on the limit of a composite function:
\[ \begin{aligned} \lim_{x\to-3}\sqrt[3]{2x+5} &= \sqrt[3]{\lim_{x\to-3}(2x+5)}\\ &= \sqrt[3]{-1}\\ &=-1. \end{aligned} \]
The result is
\[ -1. \]
Exercise 8 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{x\to 0}\cos\left(\frac{x^2+1}{x+1}\right). \]
Result
\[ \lim_{x\to 0}\cos\left(\frac{x^2+1}{x+1}\right)=\cos(1). \]
Solution
The expression is a composite function. The inner function is
\[ f(x)=\frac{x^2+1}{x+1}, \]
while the outer function is
\[ \varphi(t)=\cos(t). \]
Let us first compute the limit of the inner function.
The numerator tends to
\[ \lim_{x\to 0}(x^2+1)=0^2+1=1. \]
The denominator tends to
\[ \lim_{x\to 0}(x+1)=1. \]
Since the limit of the denominator is non-zero, we may apply the quotient rule:
\[ \lim_{x\to 0}\frac{x^2+1}{x+1} = \frac{1}{1} = 1. \]
The cosine function is continuous at every real point. Applying the theorem on the limit of a composite function,
\[ \begin{aligned} \lim_{x\to 0}\cos\left(\frac{x^2+1}{x+1}\right) &= \cos\left( \lim_{x\to 0}\frac{x^2+1}{x+1} \right)\\ &=\cos(1). \end{aligned} \]
Hence the required limit is
\[ \cos(1). \]
Exercise 9 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to 3}\frac{x^2-9}{x-3}. \]
Result
\[ \lim_{x\to 3}\frac{x^2-9}{x-3}=6. \]
Solution
If we substitute \(x=3\) directly, the numerator and denominator both vanish:
\[ \frac{3^2-9}{3-3} = \frac{0}{0}. \]
The symbol \(\displaystyle\frac{0}{0}\) is an indeterminate form. It does not represent the value of the limit, but merely indicates that the quotient rule cannot be applied directly.
We must transform the expression. The numerator is a difference of squares:
\[ x^2-9=x^2-3^2=(x-3)(x+3). \]
For \(x\neq 3\), we may therefore write
\[ \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x+3. \]
This simplification is legitimate in the computation of the limit, since we consider values of \(x\) close to \(3\) but distinct from \(3\).
The two functions
\[ \frac{x^2-9}{x-3} \qquad\text{and}\qquad x+3 \]
coincide, in fact, at every point of the domain with \(x\neq 3\). They therefore have the same limit as \(x\to 3\).
We obtain
\[ \lim_{x\to 3}\frac{x^2-9}{x-3} = \lim_{x\to 3}(x+3) = 3+3 = 6. \]
Hence
\[ 6. \]
Exercise 10 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to 1}\frac{x^3-1}{x^2-1}. \]
Result
\[ \lim_{x\to 1}\frac{x^3-1}{x^2-1} = \frac{3}{2}. \]
Solution
Direct substitution gives
\[ \frac{1^3-1}{1^2-1} = \frac{0}{0}. \]
We are thus faced with an indeterminate form. Let us look for a common factor of the numerator and denominator.
The numerator is a difference of cubes:
\[ x^3-1^3=(x-1)(x^2+x+1). \]
The denominator is a difference of squares:
\[ x^2-1^2=(x-1)(x+1). \]
For \(x\neq 1\), we may therefore cancel the factor \(x-1\):
\[ \begin{aligned} \frac{x^3-1}{x^2-1} &= \frac{(x-1)(x^2+x+1)}{(x-1)(x+1)}\\ &= \frac{x^2+x+1}{x+1}. \end{aligned} \]
In the new expression the denominator tends to
\[ 1+1=2\neq 0. \]
We may therefore apply the quotient rule:
\[ \begin{aligned} \lim_{x\to 1}\frac{x^3-1}{x^2-1} &= \lim_{x\to 1}\frac{x^2+x+1}{x+1}\\ &= \frac{1^2+1+1}{1+1}\\ &= \frac{3}{2}. \end{aligned} \]
The result is
\[ \frac{3}{2}. \]
Exercise 11 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x}. \]
Result
\[ \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x} = \frac{1}{2}. \]
Solution
For \(x\) sufficiently close to \(0\), we have \(1+x>0\), so the square root is defined.
Substituting \(x=0\) directly, we obtain
\[ \frac{\sqrt{1+0}-1}{0} = \frac{1-1}{0} = \frac{0}{0}. \]
The form is indeterminate. The presence of a difference involving a square root suggests rationalising the numerator.
Let us multiply numerator and denominator by the conjugate expression \(\sqrt{1+x}+1\):
\[ \frac{\sqrt{1+x}-1}{x} \cdot \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}. \]
In the numerator we use the product of a sum and a difference:
\[ \bigl(\sqrt{1+x}-1\bigr)\bigl(\sqrt{1+x}+1\bigr) = (1+x)-1 = x. \]
Hence, for \(x\neq 0\),
\[ \begin{aligned} \frac{\sqrt{1+x}-1}{x} &= \frac{x}{x\bigl(\sqrt{1+x}+1\bigr)}\\ &= \frac{1}{\sqrt{1+x}+1}. \end{aligned} \]
The new expression no longer displays any indeterminate form. Indeed, the denominator tends to
\[ \sqrt{1+0}+1=2. \]
Applying the quotient rule,
\[ \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x} = \lim_{x\to 0}\frac{1}{\sqrt{1+x}+1} = \frac{1}{2}. \]
Hence
\[ \frac{1}{2}. \]
Exercise 12 — level ★★★☆☆
Study the limit
\[ \lim_{x\to 2}\frac{x+1}{x-2}. \]
Result
\[ \lim_{x\to 2^-}\frac{x+1}{x-2}=-\infty, \qquad \lim_{x\to 2^+}\frac{x+1}{x-2}=+\infty. \]
Hence the two-sided limit as \(x\to 2\) does not exist, even in the extended real sense.
Solution
The numerator tends to
\[ \lim_{x\to 2}(x+1)=3, \]
while the denominator tends to
\[ \lim_{x\to 2}(x-2)=0. \]
We cannot apply the quotient rule, since the limit of the denominator is zero.
Since the numerator tends to the positive number \(3\), by the sign-preservation property of limits it is positive for every \(x\) sufficiently close to \(2\).
The sign of the quotient therefore depends on the sign of the denominator \(x-2\).
Let us first consider the left-hand limit. As \(x\to 2^-\), we have \(x<2\), hence
\[ x-2<0. \]
Moreover, the denominator approaches zero through negative values:
\[ x-2\to 0^-. \]
The quotient of a positive quantity close to \(3\) by a negative quantity of arbitrarily small absolute value assumes negative values of arbitrarily large modulus. Hence,
\[ \lim_{x\to 2^-}\frac{x+1}{x-2}=-\infty. \]
Let us now consider the right-hand limit. As \(x\to 2^+\), we have \(x>2\), hence
\[ x-2>0 \]
and
\[ x-2\to 0^+. \]
The quotient of a positive quantity close to \(3\) by a positive quantity tending to zero becomes arbitrarily large. It follows that
\[ \lim_{x\to 2^+}\frac{x+1}{x-2}=+\infty. \]
Since the left-hand and right-hand limits differ, the two-sided limit does not exist:
\[ \lim_{x\to 2}\frac{x+1}{x-2}\text{ does not exist}. \]
Exercise 13 — level ★★★★☆
Evaluate the limit
\[ \lim_{x\to+\infty}x\cdot\frac{1}{x+1}. \]
Result
\[ \lim_{x\to+\infty}x\cdot\frac{1}{x+1}=1. \]
Solution
As \(x\to+\infty\), the first factor tends to \(+\infty\):
\[ x\to+\infty. \]
The second factor, on the other hand, tends to zero:
\[ \frac{1}{x+1}\to 0. \]
The product thus has the indeterminate form
\[ \infty\cdot 0. \]
We cannot conclude that the result is zero or infinite. We must transform the expression.
Let us write the product as a single quotient:
\[ x\cdot\frac{1}{x+1} = \frac{x}{x+1}. \]
Numerator and denominator both tend to \(+\infty\), so the indeterminate form \(\displaystyle\frac{\infty}{\infty}\) appears.
Let us divide numerator and denominator by \(x\), which is non-zero for every sufficiently large \(x\):
\[ \frac{x}{x+1} = \frac{\displaystyle\frac{x}{x}} {\displaystyle\frac{x+1}{x}} = \frac{1}{1+\displaystyle\frac{1}{x}}. \]
Now
\[ \frac{1}{x}\to 0. \]
The denominator of the new expression therefore tends to
\[ 1+0=1, \]
which is non-zero. We may apply the quotient rule:
\[ \lim_{x\to+\infty}\frac{1}{1+\displaystyle\frac{1}{x}} = \frac{1}{1} = 1. \]
Hence
\[ 1. \]
Exercise 14 — level ★★★★☆
Evaluate the limit
\[ \lim_{x\to+\infty}\left(\sqrt{x^2+4x}-x\right). \]
Result
\[ \lim_{x\to+\infty}\left(\sqrt{x^2+4x}-x\right)=2. \]
Solution
As \(x\to+\infty\), we have
\[ x^2+4x\to+\infty, \]
and hence
\[ \sqrt{x^2+4x}\to+\infty. \]
The second term also tends to \(+\infty\). The difference therefore has the indeterminate form
\[ \infty-\infty. \]
To resolve this indeterminate form, we rationalise the expression by multiplying and dividing by its conjugate:
\[ \sqrt{x^2+4x}+x. \]
We obtain
\[ \sqrt{x^2+4x}-x = \frac{\bigl(\sqrt{x^2+4x}-x\bigr)\bigl(\sqrt{x^2+4x}+x\bigr)} {\sqrt{x^2+4x}+x}. \]
In the numerator a difference of squares appears:
\[ \bigl(\sqrt{x^2+4x}\bigr)^2-x^2 = x^2+4x-x^2 = 4x. \]
Hence,
\[ \sqrt{x^2+4x}-x = \frac{4x}{\sqrt{x^2+4x}+x}. \]
Since \(x\to+\infty\), we have \(x>0\) for all sufficiently large \(x\). Hence \(\sqrt{x^2}=|x|=x\), and
\[ \sqrt{x^2+4x} = \sqrt{x^2\left(1+\frac{4}{x}\right)} = x\sqrt{1+\frac{4}{x}}. \]
Substituting,
\[ \frac{4x}{\sqrt{x^2+4x}+x} = \frac{4x}{x\sqrt{1+\displaystyle\frac{4}{x}}+x}. \]
Let us factor \(x\) out of the denominator:
\[ \frac{4x}{x\left(\sqrt{1+\displaystyle\frac{4}{x}}+1\right)} = \frac{4}{\sqrt{1+\displaystyle\frac{4}{x}}+1}. \]
Since
\[ \frac{4}{x}\to 0, \]
it follows that
\[ \sqrt{1+\frac{4}{x}}\to 1. \]
The limit thus becomes
\[ \frac{4}{1+1} = \frac{4}{2} = 2. \]
Hence
\[ 2. \]
Exercise 15 — level ★★★★☆
Evaluate the limit
\[ \lim_{x\to-\infty}\left(-2x^5+3x^2-1\right). \]
Result
\[ \lim_{x\to-\infty}\left(-2x^5+3x^2-1\right)=+\infty. \]
Solution
The polynomial contains terms which, considered separately, behave differently:
\[ -2x^5\to+\infty, \qquad 3x^2\to+\infty, \qquad -1\to-1. \]
The result can be determined by observing that the behaviour of a polynomial at infinity is governed by its term of highest degree.
The dominant term is
\[ -2x^5. \]
To make the comparison rigorous, let us factor out \(x^5\):
\[ -2x^5+3x^2-1 = x^5\left(-2+\frac{3}{x^3}-\frac{1}{x^5}\right). \]
When \(x\to-\infty\), the power \(x^5\), having odd exponent, tends to \(-\infty\):
\[ x^5\to-\infty. \]
Moreover,
\[ \frac{3}{x^3}\to 0 \qquad\text{and}\qquad \frac{1}{x^5}\to 0. \]
Hence the second factor tends to
\[ -2+0-0=-2. \]
This limit is negative and non-zero. By the sign-preservation property of limits, the factor in brackets is negative for every sufficiently negative \(x\).
The product therefore consists of a negative quantity of arbitrarily large absolute value and a negative quantity close to \(-2\). It is positive and becomes arbitrarily large:
\[ (-\infty)\cdot(-2)=+\infty. \]
This notation is merely shorthand for the corresponding statement about limits. We conclude that
\[ +\infty. \]
Exercise 16 — level ★★★★☆
Evaluate
\[ \lim_{x\to\pm\infty}\frac{3x^2-1}{x^4+2}. \]
Result
\[ \lim_{x\to+\infty}\frac{3x^2-1}{x^4+2}=0, \qquad \lim_{x\to-\infty}\frac{3x^2-1}{x^4+2}=0. \]
Solution
The numerator is a polynomial of degree \(2\), while the denominator is a polynomial of degree \(4\).
As \(x\to+\infty\) and as \(x\to-\infty\), both the numerator and the denominator tend to \(+\infty\). The symbolic substitution therefore yields the indeterminate form
\[ \frac{\infty}{\infty}. \]
Since the degree of the denominator exceeds that of the numerator, we expect the denominator to grow faster and the quotient to tend to zero.
Let us divide numerator and denominator by \(x^4\), the highest power appearing in the denominator:
\[ \frac{3x^2-1}{x^4+2} = \frac{\displaystyle\frac{3x^2}{x^4}-\displaystyle\frac{1}{x^4}} {\displaystyle\frac{x^4}{x^4}+\displaystyle\frac{2}{x^4}}. \]
Simplifying the powers,
\[ \frac{3x^2-1}{x^4+2} = \frac{\displaystyle\frac{3}{x^2}-\displaystyle\frac{1}{x^4}} {1+\displaystyle\frac{2}{x^4}}. \]
As \(x\to+\infty\) and as \(x\to-\infty\),
\[ \frac{1}{x^2}\to 0 \qquad\text{and}\qquad \frac{1}{x^4}\to 0. \]
The numerator of the new fraction therefore tends to
\[ 3\cdot 0-0=0, \]
while the denominator tends to
\[ 1+0=1. \]
Since the limit of the denominator is non-zero, we may apply the quotient rule:
\[ \lim_{x\to\pm\infty}\frac{3x^2-1}{x^4+2} = \frac{0}{1} = 0. \]
Hence, in both directions,
\[ 0. \]
Exercise 17 — level ★★★★★
Evaluate the limit
\[ \lim_{x\to-\infty}\frac{2x^3-x+1}{4x^3+5}. \]
Result
\[ \lim_{x\to-\infty}\frac{2x^3-x+1}{4x^3+5} = \frac{1}{2}. \]
Solution
The numerator and denominator are polynomials of the same degree, namely degree \(3\).
As \(x\to-\infty\),
\[ 2x^3-x+1\to-\infty \]
and
\[ 4x^3+5\to-\infty. \]
The quotient thus has the indeterminate form
\[ \frac{\infty}{\infty}. \]
Let us divide every term by \(x^3\), the power corresponding to the highest degree:
\[ \frac{2x^3-x+1}{4x^3+5} = \frac{\displaystyle\frac{2x^3}{x^3} -\displaystyle\frac{x}{x^3} +\displaystyle\frac{1}{x^3}} {\displaystyle\frac{4x^3}{x^3} +\displaystyle\frac{5}{x^3}}. \]
Simplifying,
\[ \frac{2x^3-x+1}{4x^3+5} = \frac{2-\displaystyle\frac{1}{x^2} +\displaystyle\frac{1}{x^3}} {4+\displaystyle\frac{5}{x^3}}. \]
As \(x\to-\infty\),
\[ \frac{1}{x^2}\to 0 \qquad\text{and}\qquad \frac{1}{x^3}\to 0. \]
The numerator therefore tends to
\[ 2-0+0=2, \]
while the denominator tends to
\[ 4+0=4. \]
The limit of the denominator is non-zero. Applying the quotient rule,
\[ \lim_{x\to-\infty}\frac{2x^3-x+1}{4x^3+5} = \frac{2}{4} = \frac{1}{2}. \]
The result coincides with the ratio of the coefficients of the terms of highest degree:
\[ \frac{2}{4}=\frac{1}{2}. \]
Hence
\[ \frac{1}{2}. \]
Exercise 18 — level ★★★★★
Evaluate the limit
\[ \lim_{x\to-\infty}\frac{x^4+1}{-2x^2+3}. \]
Result
\[ \lim_{x\to-\infty}\frac{x^4+1}{-2x^2+3} = -\infty. \]
Solution
The numerator is a polynomial of degree \(4\), while the denominator is a polynomial of degree \(2\).
As \(x\to-\infty\), both powers \(x^4\) and \(x^2\) tend to \(+\infty\). It follows that
\[ x^4+1\to+\infty \]
and
\[ -2x^2+3\to-\infty. \]
To determine rigorously the behaviour of the quotient, let us divide numerator and denominator by \(x^2\):
\[ \frac{x^4+1}{-2x^2+3} = \frac{\displaystyle\frac{x^4}{x^2} +\displaystyle\frac{1}{x^2}} {\displaystyle\frac{-2x^2}{x^2} +\displaystyle\frac{3}{x^2}}. \]
Simplifying,
\[ \frac{x^4+1}{-2x^2+3} = \frac{x^2+\displaystyle\frac{1}{x^2}} {-2+\displaystyle\frac{3}{x^2}}. \]
As \(x\to-\infty\),
\[ x^2\to+\infty \qquad\text{and}\qquad \frac{1}{x^2}\to 0. \]
The numerator therefore tends to \(+\infty\):
\[ x^2+\frac{1}{x^2}\to+\infty. \]
The denominator, on the other hand, tends to the negative number
\[ -2+0=-2. \]
By the sign-preservation property of limits, the denominator is negative and stays bounded away from zero for every sufficiently negative \(x\).
Dividing a positive quantity of arbitrarily large size by a negative quantity close to \(-2\) yields a negative quantity of arbitrarily large absolute value.
Hence
\[ -\infty. \]
Exercise 19 — level ★★★★★
Evaluate the limit
\[ \lim_{x\to+\infty}x^{\frac{1}{\ln(x)}}. \]
Result
\[ \lim_{x\to+\infty}x^{\frac{1}{\ln(x)}}=e. \]
Solution
The function is defined for \(x>0\) and \(x\neq 1\), since the natural logarithm requires \(x>0\) and the denominator of the exponent cannot equal zero.
As \(x\to+\infty\), the base tends to \(+\infty\):
\[ x\to+\infty. \]
Since
\[ \ln(x)\to+\infty, \]
the exponent tends to zero:
\[ \frac{1}{\ln(x)}\to 0. \]
The power thus has the indeterminate form
\[ \infty^0. \]
We must not, however, try to assign a value to the symbolic form itself. In this case the expression can be simplified exactly, using the relation between the exponential function and the natural logarithm.
For every \(x>0\),
\[ x=e^{\ln(x)}. \]
Hence, for \(x>1\),
\[ x^{\frac{1}{\ln(x)}} = \left(e^{\ln(x)}\right)^{\frac{1}{\ln(x)}}. \]
Applying the property \((a^b)^c=a^{bc}\), which is legitimate here since the base \(e\) is positive, we obtain
\[ \left(e^{\ln(x)}\right)^{\frac{1}{\ln(x)}} = e^{\ln(x)\cdot\frac{1}{\ln(x)}}. \]
Since \(x\neq 1\), we have \(\ln(x)\neq 0\), and so
\[ \ln(x)\cdot\frac{1}{\ln(x)}=1. \]
It follows that, for every \(x>1\),
\[ x^{\frac{1}{\ln(x)}}=e. \]
The function is therefore constant and equal to \(e\) throughout the interval \((1,+\infty)\). Its limit is accordingly
\[ e. \]
Exercise 20 — level ★★★★★
Study the limit
\[ \lim_{x\to 1} \frac{(x-1)^2(x+2)} {(x-1)^3(2-x)}. \]
Result
\[ \lim_{x\to 1^-} \frac{(x-1)^2(x+2)} {(x-1)^3(2-x)} = -\infty, \]
\[ \lim_{x\to 1^+} \frac{(x-1)^2(x+2)} {(x-1)^3(2-x)} = +\infty. \]
Hence the two-sided limit as \(x\to 1\) does not exist, even in the extended real sense.
Solution
Substituting \(x=1\) directly, both the numerator and the denominator tend to zero:
\[ (x-1)^2(x+2)\to 0 \]
and
\[ (x-1)^3(2-x)\to 0. \]
The initial form is therefore
\[ \frac{0}{0}. \]
The factor \(x-1\) appears in both numerator and denominator, but with different multiplicities. In the numerator it has exponent \(2\), while in the denominator it has exponent \(3\).
For \(x\neq 1\), we may therefore cancel the common factor \((x-1)^2\):
\[ \begin{aligned} \frac{(x-1)^2(x+2)} {(x-1)^3(2-x)} &= \frac{(x-1)^2(x+2)} {(x-1)^2(x-1)(2-x)}\\ &= \frac{x+2}{(x-1)(2-x)}. \end{aligned} \]
After the simplification, a factor \(x-1\) remains in the denominator. This indicates that the absolute value of the quotient will tend to infinity, but to establish its sign we must examine each factor separately.
As \(x\to 1\),
\[ x+2\to 3. \]
Since \(3>0\), by the sign-preservation property of limits the factor \(x+2\) is positive in a suitable neighbourhood of \(1\).
Furthermore,
\[ 2-x\to 1. \]
This limit, too, is positive, so \(2-x\) is positive for every \(x\) sufficiently close to \(1\).
The only factor that changes sign as \(x\) crosses \(1\) is therefore \(x-1\).
Let us consider the left-hand limit. If \(x\to 1^-\), then \(x<1\), so
\[ x-1<0 \]
and
\[ x-1\to 0^-. \]
The numerator \(x+2\) is positive. In the denominator, the factor \(x-1\) is negative while the factor \(2-x\) is positive. The denominator is thus negative and tends to zero.
It follows that
\[ \lim_{x\to 1^-} \frac{x+2}{(x-1)(2-x)} = -\infty. \]
Let us now consider the right-hand limit. If \(x\to 1^+\), then \(x>1\), so
\[ x-1>0 \]
and
\[ x-1\to 0^+. \]
All the factors \(x+2\), \(x-1\) and \(2-x\) are positive for \(x\) sufficiently close to \(1\) from the right. The denominator tends to zero through positive values.
Hence,
\[ \lim_{x\to 1^+} \frac{x+2}{(x-1)(2-x)} = +\infty. \]
The two one-sided limits differ:
\[ -\infty\neq+\infty. \]
Consequently,
\[ \lim_{x\to 1} \frac{(x-1)^2(x+2)} {(x-1)^3(2-x)} \text{ does not exist}. \]