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

From UBC Wiki
Jump to: navigation, search

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 the definition of derivative, first consider the right hand limit:

The last inequality follows from

Then, the left hand limit can be computed as

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