Differentiability of an even function If $f$ is an even function such that $\lim_{h \to 0} \frac{f(h)-f(0)}{h}$ has a finite non zero value , then is $f(x)$ continuous , differentiable , or neither continuous nor differentiable at $x=0$? 
I think that the function is continuous , and it is easy to prove that too. But it would not be differentiable. This is because , the above expression represents the derivative of the function at $x=0$ and and the derivative of an even function at $0$ is always zero however according to the given question , the derivative of the function is not zero. So it is not differentiable. Do you think I have reasoned it correctly ?

 A: By the Principle of Explosion, all answers are correct.  There are two contradictory statements in the question:


*

*It is stated that $f'(0)$ exists (because the limit of the difference quotient exists) and that $f$ is even.  Hence
\begin{align} f'(0) = \lim_{h\to 0} \frac{f(h)-f(0)}{h}
&= \lim_{h\to 0} \frac{f(-h)-f(0)}{h} \\
&= \lim_{-h\to 0} \frac{f(h)-f(0)}{-h} \\
&= -\lim_{h\to 0} \frac{f(h)-f(0)}{h} \\
&= -f'(0).
\end{align}
Since $f'(0) = -f'(0)$, it must be the case that $f'(0) = 0$.

*It is also state that $f'(0)$ exists and $f'(0) > 0$.


These two statements are contradictory, therefore any conclusion follows.  The correct answer, then, is to mark all of the multiple choice options. ;)

While it is impossible to know, it is likely that (as has been pointed out in the comments) the author(s) intended to consider the one sided limit.  It is reasonable to conclude that the question should have read

If $f$ is an even function such that
  $$ \lim_{h\to 0^+} \frac{f(h)-f(0)}{h} $$
  has some finite, non-zero value, then... (multiple choice options).

In this case, there is no contradiction.  Then, by the reasoning above, we know that the function cannot be differentiable at $x=0$.  If it were, then the derivative would be zero, but we know that it is not.
On the other hand, suppose that the one-sided limit is $L$, i.e. that $\lim_{h\to 0^+} (f(h)-f(0))/h = L$.  Then
$$ |f(h) - f(0)|
= \left| h \frac{f(h) - f(0)}{h} \right|
= |h| \left|  \frac{f(h) - f(0)}{h} \right|
= |h| |L|. $$
We can make this as small as we like by choosing $h$ sufficiently small.  The limit from the left will be the same (the sign of the limits will differ, but this is absorbed into the absolute value).  This implies that $f$ is continuous at zero.  Therefore, assuming that the question was supposed to read as indicated above, the correct answer would be (b).
A: Well, the given question isn't really valid, because it's not possible for $f$ to be an even function such that $\lim_{h \to 0} \frac{f(h)-f(0)}{h}$ has a finite non zero value.
Your reasoning isn't really correct, because you should have concluded that the given conditions are impossible. "The function is not differentiable at $0$" is not the correct conclusion to come to.
In my opinion, none of the given answer choices are accurate.
A: So I think the answer is that the you are right: function is continuous but not differentiable.
I am assuming that the question meant 'limit from the positive direction' otherwise, as others have pointed out, the question is contradictory. However this is a reasonable assumption because with this assumption the question becomes meaningful and interesting.
The limit 'from the right; the positive side' is non-zero. The function is even and so the limit that comes from the left (the negative side) will have the opposite sign. And so these two limits cannot be equal (because they are non-zero) which means that the function does not have a derivative. For a mental picture, imagine two straight lines (one from the top right and one from the top left) into the origin and forming a point. At the origin there is no derivative.
But the function is continuous essentially because we can make the distance between $f(h)$ and $f(0)$ as small we we please by making $h$ small and then the difference between $f(h)$ and $f(0)$ will look like $h.l$ where $l$ is the limit of $(f(h)-f(0))/h$. We will be able to do something similar (with reversed sign) from the other side because $f$ is even. So, yes, I think you are right: $f$ is continuous but not differentiable.
