# L'Hospital's Rule

L'Hospital's Rule is useful when determining the behavior of a function having a limit of indeterminate form. This discussion will focus on two types of indeterminate forms.

The first type occurs when the limit of a function results in a limit of $\frac{0}{0}$ and is said to be a limit of indeterminate form $\frac{0}{0}$.

INDETERMINATE FORM $\frac{0}{0}$:

$\underset{x\to 1}{\mathrm{lim}}\frac{\mathrm{log}x}{x-1}=\frac{\mathrm{log}1}{1-1}=\frac{0}{0}$

The second type occurs when the limit of a function results in a limit of $\frac{\infty }{\infty }$ and is said to be a limit of indeterminate form $\frac{\infty }{\infty }$.

INDETERMINATE FORM $\frac{\infty }{\infty }$:

$\underset{x\to \infty }{\mathrm{lim}}\frac{\mathrm{log}x}{x-1}=\frac{\mathrm{log}\infty }{\infty -1}=\frac{\infty }{\infty }$

Let f(x) and g(x) be two functions that are differentiable on an open interval with point s, except maybe at s, and the first derivative of g(x) is not zero. If the limit of $\frac{f\left(x\right)}{g\left(x\right)}$ as x approaches s is of indeterminate form $\frac{0}{0}$ or $\frac{\infty }{\infty }$ then L'Hospital's Rule can be applied.

L'Hospital's Rule states that the limit of the quotient of the two functions is equal to the limit of the quotients of their first derivatives.

L'HOSPITAL'S RULE:

If f(x) and g(x) are differentiable on an open interval that contains s (except maybe at s) and the limit is of indeterminate form $\frac{0}{0}$ or $\frac{\infty }{\infty }$, that is:

$\underset{x\to s}{\mathrm{lim}}f\left(x\right)=0$AND$\underset{x\to s}{\mathrm{lim}}g\left(x\right)=0$

OR

$\underset{x\to \infty }{\mathrm{lim}}f\left(x\right)=\infty$     AND$\underset{x\to \infty }{\mathrm{lim}}g\left(x\right)=\infty$

Then the following rule applies:

$\underset{x\to s}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to s}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}$

Let's look at some examples.

To work these examples requires the use of various derivative rules. If you are not familiar with a rule go to the associated topic for a review.

Example 1: Find $\underset{x\to 2}{\mathrm{lim}}\frac{{x}^{2}-2x}{{x}^{3}+x-10}$.
 Step 1: Ensure that the limit is of indeterminate form $\frac{0}{0}$ or $\frac{\infty }{\infty }$. Take the limit: $\underset{x\to 2}{\mathrm{lim}}\frac{{x}^{2}-2x}{{x}^{3}+x-10}$ $\underset{x\to 2}{\mathrm{lim}}\frac{{\left(2\right)}^{2}-2\left(2\right)}{{\left(2\right)}^{3}+\left(2\right)-10}=\frac{0}{0}$ Step 2: Apply L'Hospital's Rule Because the limit is of indeterminate form $\frac{0}{0}$, L'Hospital's rule can be applied. $\underset{x\to s}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to s}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}$ $\underset{x\to 2}{\mathrm{lim}}\frac{{x}^{2}-2x}{{x}^{3}+x-10}$ $\underset{x\to 2}{\mathrm{lim}}\frac{2x-2}{3{x}^{2}+1}$ Take the derivative $\underset{x\to 2}{\mathrm{lim}}\frac{2\left(2\right)-2}{3{\left(2\right)}^{2}+1}=\frac{2}{13}$ Take the limit
Example 1: Find $\underset{x\to \infty }{\mathrm{lim}}\frac{\mathrm{ln}x}{{x}^{3}+1}$.
 Step 1: Ensure that the limit is of indeterminate form $\frac{0}{0}$ or $\frac{\infty }{\infty }$. Take the limit: $\underset{x\to \infty }{\mathrm{lim}}\frac{\mathrm{ln}x}{{x}^{3}+1}$ $\underset{x\to \infty }{\mathrm{lim}}\frac{\mathrm{ln}x\infty }{{\left(\infty \right)}^{3}+1}=\frac{\infty }{\infty }$ Step 2: Apply L'Hospital's Rule Because the limit is of indeterminate form $\frac{\infty }{\infty }$, L'Hospital's rule can be applied. $\underset{x\to s}{\mathrm{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to s}{\mathrm{lim}}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}$ $\underset{x\to \infty }{\mathrm{lim}}\frac{\mathrm{ln}x}{{x}^{3}+1}$ $\underset{x\to \infty }{\mathrm{lim}}\frac{1/x}{3{x}^{2}}=\frac{1}{3{x}^{3}}$    Take the derivative $\underset{x\to \infty }{\mathrm{lim}}\frac{1}{3{x}^{3}}=\frac{1}{3{\left(\infty \right)}^{3}}=0$    Take the limit

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