# Solve \lim \limits_{\to 0} \frac {(x)sin(x)}{1-cos(x)}

## Question:

Solve {eq}\lim \limits_{x\to 0} \frac {x \ sin(x)}{1-cos(x)} {/eq}

## Limits of Functions:

The L-Hopitals rule when is applied to the limit expression of the equation containing the limits will get the simplified expression whose limits are found after direct replacement or we will have to apply the L-Hopitals rule again.

So to find the limit of

{eq}\lim_{x\to 0} \frac{x \ sin(x)}{1-cos(x)}\\ {/eq}

we have to do the following. Applying the L-Hopitals rule will give:

{eq}\lim _{x\to \:0}\left(\frac{\left(x\sin \left(x\right)\right)^{'\:}}{\left(1-\cos \left(x\right)\right)^{'\:}}\right)\\ =\lim _{x\to \:0}\left(\frac{\sin \left(x\right)+x\cos \left(x\right)}{\sin \left(x\right)}\right)\\ {/eq}

Again we have to apply the L-Hopitals rule here as follows:

{eq}\lim _{x\to \:0}\left(\frac{\left(\sin \left(x\right)+x\cos \left(x\right)\right)^{'\:}}{\left(\sin \left(x\right)\right)^{'\:}}\right)\\ =\lim _{x\to \:0}\left(\frac{2\cos \left(x\right)-x\sin \left(x\right)}{\cos \left(x\right)}\right)\\ =\frac{2\cos \left(0\right)-0\cdot \sin \left(0\right)}{\cos \left(0\right)}\\ =2 {/eq}

Now this is the required limit. 