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# Calculus

Limits and Continuity of Functions

# L’Hopital’s Rule

Page 1
Problems 1-3
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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
$f{\left( x \right)^{g\left( x \right)}} = {e^{g\left( x \right)\ln f\left( x \right)}}.$

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