# Science:Math Exam Resources/Courses/MATH100/December 2015/Question 07/Solution 1

From UBC Wiki

First, we check that is continuous at 0. (Note that if is not continuous at 0, is not differentiable at 0.)

Using , we can compute the right hand limit as

On other hand, it is easy to get

Therefore, is continuous at 0.

Now, we show is differentiable at 0.

By its definition, first consider the left hand limit:

The last inequality follows from

Then, the right hand limit can be computed as

Since the right hand limit matches with the left hand limit, the limit exists so that is differentiable at 0 and .