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


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.

Answer and Explanation:

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.

Learn more about this topic:

How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 6 / Lesson 4

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