Prove that $\lim_{x\rightarrow a^-} f'(x)$ and $\lim_{x\rightarrow a^+} f'(x)$ cannot both exist. Suppose that $f$ is differentiable in some interval contain $a$ but that $f'$ is discontinous at $a$. Prove that $\lim_{x\rightarrow a^-} f'(x)$ and $\lim_{x\rightarrow a^+} f'(x)$ cannot  both exist.

Attempt: I think the question should have instead been phrased as : prove that $\lim_{x\rightarrow a^-} f'(x)$ and $\lim_{x\rightarrow a^+} f'(x)$ both cannot be equal and finite

If $\lim_{x\rightarrow a^-} f'(x)=l$ and $\lim_{x\rightarrow a^+} f'(x)=m~|~l \ne m$, then $f'$ is still discontinuous at $a$ because the limit itself doesn't exist. 
On the other hand, if $\lim_{x\rightarrow a^-} f'(x)=p=\lim_{x\rightarrow a^+} f'(x)$, then, $f'(a)=p$ itself.
Please tell me what do you think. Is the question correctly phrased?

If any of the one sided limit exists and other doesn't, then won't the function cease to be differentiable? 

 A: There is a big difference between saying that the two one-sided limits of $f$' both cannot exist. or cannot both exist.


*

*both cannot exist means that the left-hand limit cannot exist, and the right-hand limit cannot exist.

*cannot both exist means it's impossible that the left-hand limit and the right-hand limit to both exist.  By De Morgan's law, it's equivalent to saying that either the left-hand limit cannot exist, or the right-hand limit cannot exist (or both).
The first statement (both cannot exist) is false.  Consider
$$
    f(x) = \begin{cases} 0 & x \leq 0 \\
                        x^2 \sin\left(\frac{1}{x}\right) & x \geq 0
\end{cases}
$$
Then $f$ is differentiable on $\mathbb R$, $f'$ is not continuous at $0$, but $\lim_{x\to 0^-} f'(x) = 0$.  The other one-sided limit $\lim_{x\to 0^+} f'(x)$ does not exist.
After comments, you clarified that the second statement (cannot both exist) is the intended one.  In that case,  I think the question is correctly phrased.  The contrapositive statement might make more sense:

Suppose that $f$ is differentiable near $a$.  If $\lim_{x\to a^-} f'(x)$ and $\lim_{x\to a^+} f'(x)$ exist [and are finite], then $f'$ is continuous at $a$.

I put and are finite in brackets because it's assumed that when a limit “exists” we're talking about a finite number.  An infinite limit doesn't “exist.”  This is complicated because we use (some might say abuse) the equals sign in both cases.  We say $\lim_{x\to 0} x = 0$ and $\lim_{x\to 0^+} \frac{1}{x} = \infty$, but the first limit “exists” while the second does not.
To dig a bit deeper, you need to show that: if $f$ is differentiable near $a$, and $\lim_{x\to a^+} f'(x)$ exists, then
$$
    \lim_{x\to a^+} \frac{f(x) - f(a)}{x-a} = \lim_{x\to a^+} f'(x)
$$
and the the same thing with $x\to a^+$ replaced by $x\to a^{-}$.  For this,
the Mean Value Theorem is your friend.
