L’Hopital’s Rule provides a method for evaluating indeterminate forms of type \(\large\frac{0}{0}\normalsize\) or \(\large\frac{\infty}{\infty}\normalsize.\)
Let \(a\) be either a finite number or infinity.
- If \(\lim\limits_{x \to a} f\left( x \right) = 0\) and \(\lim\limits_{x \to a} g\left( x \right) = 0,\) then \(\lim\limits_{x \to a} {\large\frac{{f\left( x \right)}}{{g\left( x \right)}}\normalsize} = \lim\limits_{x \to a} {\large\frac{{f’\left( x \right)}}{{g’\left( x \right)}}\normalsize};\)
- If \(\lim\limits_{x \to a} f\left( x \right) = \infty\) and \(\lim\limits_{x \to a} g\left( x \right) = \infty,\) then similarly \(\lim\limits_{x \to a} {\large\frac{{f\left( x \right)}}{{g\left( x \right)}}\normalsize} = \lim\limits_{x \to a} {\large\frac{{f’\left( x \right)}}{{g’\left( x \right)}}\normalsize}.\)
This rule appeared in \(1696\) (!) in the first book on differential calculus published by French mathematician Guillaume de l’Hopital \(\left(1661- 1704\right).\)
We can apply L’Hopital’s rule to indeterminate forms of type \(0 \cdot \infty,\) \(\infty – \infty,\) \(0^0,\) \(1^{\infty},\) \(\infty^0\) as well. The first two indeterminate forms \(0 \cdot \infty\) and \(\infty – \infty\) can be reduced to the forms \(\large\frac{0}{0}\normalsize\) or \(\large\frac{\infty}{\infty}\normalsize\) using algebraic transformations. The indeterminate forms \(0^0,\) \(1^{\infty}\) and \(\infty^0\) can be reduced to the form \(0 \cdot \infty\) with the help of identity
\[f{\left( x \right)^{g\left( x \right)}} = {e^{g\left( x \right)\ln f\left( x \right)}}.\]
L’Hopital’s rule also holds for one-sided limits.
Solved Problems
Click or tap a problem to see the solution.
Example 1
Find the limit \(\lim\limits_{x \to 0} {\large\frac{{{a^x} – 1}}{x}\normalsize}.\)Example 2
Find the limit \(\lim\limits_{x \to 2} {\large\frac{{\sqrt {7 + x} – 3}}{{x – 2}}\normalsize}.\)Example 3
Calculate the limit \(\lim\limits_{x \to 2} \left( {{\large\frac{4}{{{x^2} – 4}}\normalsize} – {\large\frac{1}{{x – 2}}\normalsize}} \right).\)Example 4
Find the limit \(\lim\limits_{x \to \infty } {\large\frac{{{x^2}}}{{{2^x}}}\normalsize}.\)Example 5
Calculate the limit \(\lim\limits_{t \to \infty } \sqrt[\large t\normalsize]{{{t^3}}}.\)Example 6
Find the limit \(\lim\limits_{x \to 1} {x^{\large\frac{1}{{1 – x}}\normalsize}}.\)Example 7
Calculate the limit \(\lim\limits_{x \to 0 + } {x^x}.\)Example 8
Calculate the limit \(\lim\limits_{x \to 0 + } {\left( {\arcsin x} \right)^{2x}}.\)Example 9
Find the limit \(\lim\limits_{x \to \pi /2} {\left( {\sin x} \right)^{\tan x}}.\)Example 10
Calculate the limit \(\lim\limits_{x \to + \infty } {\large\frac{{{x^n}}}{{{e^x}}}\normalsize}.\)Example 11
Find the limit \(\lim\limits_{x \to \pi /2} {\large\frac{{\tan x}}{{\tan 3x}}\normalsize}.\)Example 12
Calculate the limit \(\lim\limits_{x \to \pi /2} {\left( {\tan x} \right)^{\cos x}}.\)Example 1.
Find the limit \(\lim\limits_{x \to 0} {\large\frac{{{a^x} – 1}}{x}\normalsize}.\)Solution.
Differentiating the numerator and denominator separately, we get
\[
{\lim\limits_{x \to 0} \frac{{{a^x} – 1}}{x} = \left[ {\frac{0}{0}} \right] }
= {\lim\limits_{x \to 0} \frac{{{{\left( {{a^x} – 1} \right)}^\prime }}}{{{{\left( x \right)}^\prime }}} }
= {\lim\limits_{x \to 0} \frac{{{a^x}\ln a}}{1} = \ln a\lim\limits_{x \to 0} {a^x} }
= {\ln a \cdot 1 = \ln a.}
\]
Example 2.
Find the limit \(\lim\limits_{x \to 2} {\large\frac{{\sqrt {7 + x} – 3}}{{x – 2}}\normalsize}.\)Solution.
Because direct substitution leads to an indeterminate form \(\large\frac{0}{0}\normalsize,\) we can use L’Hopital’s rule:
\[
{\lim\limits_{x \to 2} \frac{{\sqrt {7 + x} – 3}}{{x – 2}} = \left[ {\frac{0}{0}} \right] }
= {\lim\limits_{x \to 2} \frac{{{{\left( {\sqrt {7 + x} – 3} \right)}^\prime }}}{{{{\left( {x – 2} \right)}^\prime }}} }
= {\lim\limits_{x \to 2} \frac{{\large\frac{1}{{2\sqrt {7 + x} }}\normalsize}}{1} }
= {\frac{1}{2}\lim\limits_{x \to 2} \frac{1}{{\sqrt {7 + x} }} = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}}
\]
Example 3.
Calculate the limit \(\lim\limits_{x \to 2} \left( {{\large\frac{4}{{{x^2} – 4}}\normalsize} – {\large\frac{1}{{x – 2}}\normalsize}} \right).\)Solution.
Here we deal with an inderminate form of type \(\infty – \infty.\) After simple transformations, we have
\[
{\lim\limits_{x \to 2} \left( {\frac{4}{{{x^2} – 4}} – \frac{1}{{x – 2}}} \right) }
= {\lim\limits_{x \to 2} \frac{{4 – \left( {x + 2} \right)}}{{{x^2} – 4}} }
= {\lim\limits_{x \to 2} \frac{{2 – x}}{{{x^2} – 4}} = \left[ {\frac{0}{0}} \right] }
= {\lim\limits_{x \to 2} \frac{{{{\left( {2 – x} \right)}^\prime }}}{{{{\left( {{x^2} – 4} \right)}^\prime }}} }
= {\lim\limits_{x \to 2} \left( {\frac{{ – 1}}{{2x}}} \right) = – \frac{1}{4}.}
\]