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Example application of l'Hôpital's rule to f(x)=sin(x) and g(x)=−0.5x: the function h(x) = f(x)/g(x) is undefined at x = 0, but can be completed to a continuous function on whole ℝ by defining h(0) = f′(0)/g′(0) = −2.

L'Hôpital's rule is a rule in calculus that helps find limits using derivatives. The name of the rule comes from the mathematician Guillaume de l'Hôpital. L'Hôpital's rule states that for any function f whose limit as f approaches some point c is an indeterminate form the value of the limit of the function is equal to the derivative of the bottom and the top. It is shown as follows:

$\lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}$

Examples

Given the limit $\lim _{x\to 5}{\frac {x^{2}-25}{x-5}}$first plug in 5 to try to find the limit. Plugging in 5 will give an indeterminate form. Then L'Hôpital's rule is applied to give ${\frac {2x}{1}}$. Then simply plugging in 5 again yields the answer of 10.

The next example involves the sine function

$\lim _{x\to 0}{\frac {sin(x)}{x}}$

$sin(0)=0,0=0,{\frac {sin(0)}{0}}={\frac {0}{0}}$ plugging in 0 gives an indeterminate form

$\lim _{x\to 0}{\frac {cos(x)}{1}}$ L'Hôpital's rule is applied and the derivatives of the top and bottom are taken

${\frac {cos(0)}{1}}=1,\lim _{x\to 0}{\frac {sin(x)}{x}}=1$ Plugging in 0 to the new expression gives 1, the answer to the limit

This example involves the natural log function

$\lim _{x\to 1}{\frac {ln(x)}{x^{2}-1}}$

$ln(1)=0,1^{2}-1=0,{\frac {ln(1)}{1^{2}-1}}={\frac {0}{0}}$ Plugging in 1 to the expression gives an indeterminate form

$\lim _{x\to 1}{\frac {\frac {1}{x}}{2x}}$ L'Hôpital's rule is applied and the derivatives of the top and bottom are taken.

${\frac {1}{2x^{2}}}$ expression is simplified

$2(1)^{2}=2,lim_{x\to 1}{\frac {1}{2x^{2}}}=1/2,lim_{x\to 1}{\frac {ln(x)}{x^{2}-1}}={\frac {1}{2}}$Plugging back in 1 gives the answer to the limit as 1/2

The next example includes both sine and natural log functions

$\lim _{x\to 1}{\frac {ln(x^{2})}{sin(x-1)}}$

$ln(1^{2})=0,sin(1-1)=0,{\frac {ln(1^{2})}{sin(1-1)}}={\frac {0}{0}}$Plugging in 1 to the expression gives an indeterminate form

$\lim _{x\to 1}{\frac {\frac {2x}{x^{2}}}{cos(x-1)}}$L'Hôpital's rule is applied and the derivatives of the top and bottom are taken.

${\frac {2x}{x^{2}cos(x-1)}}$expression is simplified

$1^{2}cos(1-1)=1,2(1)=2,{\frac {2(1)}{1^{2}cos(1-1)}}=2,\lim _{x\to 1}{\frac {ln(x)}{sin(x-1)}}=2$ Plugging back in 1 gives the answer to the limit as 2

The final example involves the tangent function

$\lim _{x\to 0}{\frac {e^{x}-1}{tan(x)}}$

$e^{0}-1=0,tan(0)=0,{\frac {e^{0}-1}{tan(0)}}={\frac {0}{0}}$Plugging in 0 to the expression gives an indeterminate form

$\lim _{x\to 0}{\frac {e^{x}}{sec^{2}(x)}}$L'Hôpital's rule is applied and the derivatives of the top and bottom are taken.

$e^{0}=1,sec^{2}(0)=1,{\frac {e^{0}}{sec^{2}(0)}}=1,\lim _{x\to 0}{\frac {e^{x}-1}{tan(x)}}=1$Plugging back in 0 gives the answer to the limit as 1

See also

• L'Hôpital controversy

Category:Articles containing proofs Category:Theorems in calculus Category:Theorems in real analysis Category:Limits (mathematics)