Calculus

Limits and Continuity of Functions

L’Hopital’s Rule

Page 1
Problems 1-3
Page 2
Problems 4-12

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
\[f{\left( x \right)^{g\left( x \right)}} = {e^{g\left( x \right)\ln f\left( x \right)}}.\]

French mathematician Guillaume de l’Hopital (1661- 1704)

Fig.1 Guillaume de l’Hopital
(1661- 1704)

L’Hopital’s rule also holds for one-sided limits.

Solved Problems

Click on problem description to see 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
\[
{\lim\limits_{x \to 2} \left( {\frac{4}{{{x^2} – 4}} – \frac{1}{{x – 2}}} \right) }
= {\lim\limits_{x \to 2} \frac{{4 – \left( {x + 2} \right)}}{{{x^2} – 4}} }
= {\lim\limits_{x \to 2} \frac{{2 – x}}{{{x^2} – 4}} = \left[ {\frac{0}{0}} \right] }
= {\lim\limits_{x \to 2} \frac{{{{\left( {2 – x} \right)}^\prime }}}{{{{\left( {{x^2} – 4} \right)}^\prime }}} }
= {\lim\limits_{x \to 2} \left( {\frac{{ – 1}}{{2x}}} \right) = – \frac{1}{4}.}
\]

Page 1
Problems 1-3
Page 2
Problems 4-12