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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.

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

$$1^2cos(1-1)=1,2(1)=2,\frac{2(1)}{1^2cos(1-1)}=2,\lim_{x \to 1}\frac{ln(x)}{sin(x-1)}=2$$ 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