Limit of Derivative and Continuity

Question

Lat $x < y < z$. Suppose $f : (x, z) \rightarrow \mathbb{R}$ is continuous and differentiable at every point of $(x, z)$ \ ${y}$, and that $f'$ has a limit at $y$. Prove that $f$ is differentiable at $y$ and that $f'(y) = \lim_{x \to y} f'(x)$.

How will we go about showing differentiability and the limit in this case? Firstly, I'm unsure what is meant by the fact that "$f'$ has a limit at $y$" and how is that relevant? Moreover, am I allow to use the mean-value theorem or does that require differentiability at y?

Just apply the defintion of derivative and apply MVT for $f(y+h)-f(y)$. — Kavi Rama Murthy, Apr 10 '20 at 09:11
@HansLundmark, that link answers my question somewhat. But in my case, I have to prove the left-hand derivative as well. How can I find both the left-hand and right-hand derivatives and show that they are both equal? — throwawayAccountForRandomStuff, Apr 10 '20 at 10:56

score 1 · Answer 1 · answered Apr 10 '20 at 09:04

1

To answer to this question : what is meant by the fact that $f'$ has a limit at $y$ and how is that relevant?

It simply mean that $f'$ is $\mathcal C^1$.

A classical example is $$f(x)=\begin{cases}x^2\cos\left(\frac{1}{x}\right)&x\neq 0\\0&x=0\end{cases}.$$

This function is derivable on the whole set $\mathbb R$ but $f'$ is not $\mathcal C^1(\mathbb R)$ since $f'$ has no limit at $0$.

Hint

By Mean value theorem, for all $h$ s.t. $(y-h,y+h)\subset (x,z)$, there is $z_h$ s.t. $|z_h-y|< h$ and $$f(y+h)-f(y)=f'(z_h)h.$$ I let you conclude.

answered Apr 10 '20 at 09:04

Surb

55,662

Sorry, but what do you mean by "$\mathcal C^1$". It's the first time I've heard this. – throwawayAccountForRandomStuff Apr 10 '20 at 10:20
It means that $f$ and $f'$ are continuous. @throwawayAccountForRandomStuff – Surb Apr 10 '20 at 11:22

Limit of Derivative and Continuity

1 Answers1