If $\lim_{x \rightarrow 1} f'(x)=2015$. Prove that $f$ is differentiable at $x=1$.

Question

Let $f:[0,2] \rightarrow \mathbb R$ be a continuous function such that, $f$ is differentiable everywhere on $[0,2]$ except possibly at $x=1.$ If $\lim_{x \rightarrow 1} f~'(x)=2015$. Prove that $f$ is differentiable at $x=1$.

Attempt:

$f:[0,2] \rightarrow \mathbb R$ be a continuous function $\implies ~\exists~m,M \in \mathbb R$ such that $m \le f(x) \le M~\forall x\in [0,2] $

It is given that $\lim_{x\rightarrow 1} f~'(x) = 2015$. We need to prove that $f~'(1) $ exists or that $$ \lim_{h^+ \rightarrow 0 } \dfrac {f(1-h)-f(1)}{h} = \lim_{h^+ \rightarrow 0 } \dfrac {f(1+h)-f(1)}{h}$$ Could someone please give me a hint on how to move ahead with this problem?

Thank you for your help in this regard.

score 7 · Answer 1 · answered Jun 18 '15 at 07:47

7

Hint: write the definition of the derivative and use the Mean Value Theorem.

answered Jun 18 '15 at 07:47

Martin Argerami

205,756

1

Since the function need not be differentiable at $1,$ this means mean value theorem may not be applicable to an interval containing $1$? – MathMan Jun 18 '15 at 07:50
2

You may want to re-read the hypotheses of the MVT. – Martin Argerami Jun 18 '15 at 07:55

score 5 · Accepted Answer · answered Jun 18 '15 at 08:42

Since the function is continuous at $1$, you have that $$ \lim_{x\to1}\frac{f(x)-f(1)}{x-1} $$ is in the indeterminate form $0/0$. Thus you can apply l'Hôpital and $$ \lim_{x\to1}\frac{f(x)-f(1)}{x-1}\overset{\mathrm{(H)}}{=} \lim_{x\to1}f'(x)=2015 $$ Since the last limit exists, also the limit we started with exists.

If $\lim_{x \rightarrow 1} f'(x)=2015$. Prove that $f$ is differentiable at $x=1$.

2 Answers2