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