Examples with detailed solutions and exercises that solve limits questions related to indeterminate forms such as:
\[ \dfrac{\infty}{\infty}, \quad 0^0, \quad \infty^0, \quad 1^\infty, \quad \infty \cdot 0, \quad \infty - \infty \]In what follows, \( C \) represents the constant of integration where applicable.
Theorem
A second version of L'Hôpital's rule allows us to replace the limit problem:
\[ \dfrac{\infty}{\infty} \]with another simpler problem to solve.
If \( \lim f(x) = \infty \) and \( \lim g(x) = \infty \) and if \( \lim \dfrac{f'(x)}{g'(x)} \) has a finite value \( L \), or is of the form \( \infty \) or \( -\infty \), then:
\[ \lim \dfrac{f(x)}{g(x)} = \lim \dfrac{f'(x)}{g'(x)} \]In the above, \( \lim \) stands for \( \lim_{x \to a} f(x) \), \( \lim_{x \to a^+} f(x) \), \( \lim_{x \to a^-} f(x) \), \( \lim_{x \to \infty} f(x) \), or \( \lim_{x \to -\infty} f(x) \).
Examples with Detailed Solutions
Click on each example to view its detailed step-by-step solution.
Example 1
Find the limit:
\[ \lim_{x \to \infty} \dfrac{\ln x}{x} \]Show Solution
Since:
\[ \lim_{x \to \infty} \ln x = \infty \]and
\[ \lim_{x \to \infty} x = \infty \]we have the indeterminate form:
\[ \lim_{x \to \infty} \dfrac{\ln x}{x} = \dfrac{\infty}{\infty} \]L'Hôpital's rule can be used to evaluate the given limit question as follows:
\[ \lim_{x \to \infty} \dfrac{\ln x}{x} = \lim_{x \to \infty} \dfrac{(\ln x)'}{(x)'} \]Evaluate the derivatives in the numerator and the denominator:
\[ = \lim_{x \to \infty} \dfrac{1/x}{1} \]Evaluate the limits in the numerator and denominator:
\[ \lim_{x \to \infty} (1/x) = 0 \quad \text{and} \quad \lim_{x \to \infty} 1 = 1 \]We now evaluate the given limit:
\[ \lim_{x \to \infty} \dfrac{\ln x}{x} = \lim_{x \to \infty} \dfrac{0}{1} = 0 \]Example 2
Find the limit:
\[ \lim_{x \to \infty} x e^{-x} \]Show Solution
Note that:
\[ \lim_{x \to \infty} x = \infty \]and
\[ \lim_{x \to \infty} e^{-x} = 0 \]This is the indeterminate form \( \infty \cdot 0 \). The idea is to convert it into the indeterminate form \( \dfrac{\infty}{\infty} \) and use L'Hôpital's theorem. Note that:
\[ \lim_{x \to \infty} x e^{-x} = \lim_{x \to \infty} \dfrac{x}{e^x} = \dfrac{\infty}{\infty} \]We apply L'Hôpital's rule:
\[ = \lim_{x \to \infty} \dfrac{(x)'}{(e^x)'} = \lim_{x \to \infty} \dfrac{1}{e^x} = 0 \]Example 3
Find the limit:
\[ \lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right)^x \]Show Solution
Note that \( \lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right) = 1 \) and the above limit is given by:
\[ \lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right)^x = 1^\infty \]which is of the indeterminate form \( 1^\infty \). If we let \( t = 1 / x \), the above limit may be written as:
\[ \lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right)^x = \lim_{t \to 0^+} (1 + t)^{1/t} \]Note that as \( x \to \infty \), \( t \to 0^+ \).
Let \( y = (1 + t)^{1/t} \) and find the limit of \( \ln y \) as \( t \) approaches 0:
\[ \ln y = \ln (1 + t)^{1/t} = \dfrac{1}{t} \ln (1 + t) \]The advantage of using \( \ln y \) is that:
\[ \lim_{t \to 0} \ln y = \lim_{t \to 0} \dfrac{\ln (1 + t)}{t} = \dfrac{0}{0} \]The limit has the indeterminate form \( 0 / 0 \), and the first L'Hôpital's rule can be applied:
\[ \lim_{t \to 0} \ln y = \lim_{t \to 0} \dfrac{(\ln (1 + t))'}{(t)'} = \lim_{t \to 0} \dfrac{1/(1+t)}{1} = 1 \]Since the limit of \( \ln y = 1 \), the limit of \( y \) is \( e^1 = e \). Hence:
\[ \lim_{x \to \infty} \left(1 + \dfrac{1}{x}\right)^x = e \]Example 4
Find the limit:
\[ \lim_{x \to 0^+} \left(\dfrac{1}{x} - \dfrac{1}{\sin x}\right) \]Show Solution
Note that:
\[ \lim_{x \to 0^+} \left(\dfrac{1}{x}\right) = +\infty \]and
\[ \lim_{x \to 0^+} \left(\dfrac{1}{\sin x}\right) = +\infty \]This limit has the indeterminate form \( \infty - \infty \) and has to be converted to another form by combining terms:
\[ \lim_{x \to 0^+} \left(\dfrac{1}{x} - \dfrac{1}{\sin x}\right) = \lim_{x \to 0^+} \dfrac{\sin x - x}{x \sin x} = \dfrac{0}{0} \]We now have the indeterminate form \( 0 / 0 \) and can use L'Hôpital's theorem:
\[ \lim_{x \to 0^+} \dfrac{\sin x - x}{x \sin x} = \lim_{x \to 0^+} \dfrac{(\sin x - x)'}{(x \sin x)'} = \lim_{x \to 0^+} \dfrac{\cos x - 1}{\sin x + x \cos x} = \dfrac{0}{0} \]We have again the indeterminate form \( 0 / 0 \) and use L'Hôpital's theorem one more time:
\[ = \lim_{x \to 0^+} \dfrac{(\cos x - 1)'}{(\sin x + x \cos x)'} \] \[ = \lim_{x \to 0^+} \dfrac{-\sin x}{\cos x + \cos x - x \sin x} = \dfrac{0}{2} = 0 \]Example 5
Find the limit:
\[ \lim_{x \to 0^+} x^x \]Show Solution
We have the indeterminate form \( 0^0 \). Let \( y = x^x \) and \( \ln y = \ln (x^x) = x \ln x \). Let us now find the limit of \( \ln y \):
\[ \lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} x \ln x = 0 \cdot \infty \]The above limit has the indeterminate form \( 0 \cdot \infty \). We convert it as follows:
\[ \lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \dfrac{\ln x}{1/x} = \dfrac{-\infty}{\infty} \]It now has the indeterminate form \( \dfrac{\infty}{\infty} \) and we can use L'Hôpital's theorem:
\[ \lim_{x \to 0^+} \dfrac{\ln x}{1/x} = \lim_{x \to 0^+} \dfrac{(\ln x)'}{(1/x)'} \] \[ = \lim_{x \to 0^+} \dfrac{1/x}{-1/x^2} \] \[ = \lim_{x \to 0^+} (-x) = 0 \]The limit of \( \ln y = 0 \), so the limit of \( y = x^x \) is equal to:
\[ \lim_{x \to 0^+} x^x = e^0 = 1 \]Exercises
Calculate the following limits. Click each exercise to check your answers.
Exercise 1
\( \displaystyle \lim_{x \to \infty} (\ln x)^{1/x} \)
Show Answer
Exercise 2
\( \displaystyle \lim_{x \to \infty} (\ln x - \ln (1 + x)) \)
Show Answer
Exercise 3
\( \displaystyle \lim_{x \to \infty} \dfrac{x}{e^x} \)
Show Answer
Exercise 4
\( \displaystyle \lim_{x \to 0^+} x^{\sin x} \)