# L’Hopital’s Rule

## Trigonometry # L’Hopital’s Rule

L'Hopital's Rule provides a method for evaluating indeterminate forms of type $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}.$$

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
$\lim\limits_{x \to a} {\frac{{f\left( x \right)}}{{g\left( x \right)}}} = \lim\limits_{x \to a} {\frac{{f'\left( x \right)}}{{g'\left( x \right)}}};$
• If $$\lim\limits_{x \to a} f\left( x \right) = \infty$$ and $$\lim\limits_{x \to a} g\left( x \right) = \infty,$$ then similarly
$\lim\limits_{x \to a} {\frac{{f\left( x \right)}}{{g\left( x \right)}}} = \lim\limits_{x \to a} {\frac{{f'\left( x \right)}}{{g'\left( x \right)}}}.$

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 $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}$$ 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 $\lim\limits_{x \to 0} {\frac{{{a^x} - 1}}{x}}.$

### Example 2

Find the limit $\lim\limits_{x \to 2} {\frac{{\sqrt {7 + x} - 3}}{{x - 2}}}.$

### Example 3

Calculate the limit $\lim\limits_{x \to 2} \left( {{\frac{4}{{{x^2} - 4}}} - {\frac{1}{{x - 2}}}} \right).$

### Example 4

Find the limit $\lim\limits_{x \to \infty } {\frac{{{x^2}}}{{{2^x}}}}.$

### Example 1.

Find the limit $\lim\limits_{x \to 0} {\frac{{{a^x} - 1}}{x}}.$

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 $\lim\limits_{x \to 2} {\frac{{\sqrt {7 + x} - 3}}{{x - 2}}}.$

Solution.

Because direct substitution leads to an indeterminate form $$\frac{0}{0},$$ we can use L'Hopital's rule:

$\lim\limits_{x \to 2} \frac{{\sqrt {7 + x} - 3}}{{x - 2}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{x \to 2} \frac{{{{\left( {\sqrt {7 + x} - 3} \right)}^\prime }}}{{{{\left( {x - 2} \right)}^\prime }}} = \lim\limits_{x \to 2} \frac{{\frac{1}{{2\sqrt {7 + x} }}}}{1} = \frac{1}{2}\lim\limits_{x \to 2} \frac{1}{{\sqrt {7 + x} }} = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}.$

### Example 3.

Calculate the limit $\lim\limits_{x \to 2} \left( {{\frac{4}{{{x^2} - 4}}} - {\frac{1}{{x - 2}}}} \right).$

Solution.

Here we deal with an inderminate form of type $$\infty - \infty.$$ After simple transformations, we have

$\lim\limits_{x \to 2} \left( {\frac{4}{{{x^2} - 4}} - \frac{1}{{x - 2}}} \right) = \lim\limits_{x \to 2} \frac{{4 - \left( {x + 2} \right)}}{{{x^2} - 4}} = \lim\limits_{x \to 2} \frac{{2 - x}}{{{x^2} - 4}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{x \to 2} \frac{{{{\left( {2 - x} \right)}^\prime }}}{{{{\left( {{x^2} - 4} \right)}^\prime }}} = \lim\limits_{x \to 2} \left( {\frac{{ - 1}}{{2x}}} \right) = - \frac{1}{4}.$

### Example 4.

Find the limit $\lim\limits_{x \to \infty } {\frac{{{x^2}}}{{{2^x}}}}.$

Solution.

Using L'Hopital's rule, we can write

$\lim\limits_{x \to \infty } \frac{{{x^2}}}{{{2^x}}} = \left[ {\frac{\infty }{\infty }} \right] = \lim\limits_{x \to \infty } \frac{{{{\left( {{x^2}} \right)}^\prime }}}{{{{\left( {{2^x}} \right)}^\prime }}} = \lim\limits_{x \to \infty } \frac{{2x}}{{{2^x}\ln 2}} = \frac{2}{{\ln 2}}\lim\limits_{x \to \infty } \frac{x}{{{2^x}}} = \left[ {\frac{\infty }{\infty }} \right] = \frac{2}{{\ln 2}}\lim\limits_{x \to \infty } \frac{{{{\left( x \right)}^\prime }}}{{{{\left( {{2^x}} \right)}^\prime }}} = \frac{2}{{\ln 2}}\lim\limits_{x \to \infty } \frac{1}{{{2^x}\ln 2}} = \frac{2}{{{{\left( {\ln 2} \right)}^2}}}\lim\limits_{x \to \infty } \frac{1}{{{2^x}}} = \frac{2}{{{{\left( {\ln 2} \right)}^2}}} \cdot 0 = 0.$