1. ## Differentiable function problem

$\displaystyle f$ is a differentiable function at $\displaystyle x=1$.prove that if $\displaystyle \lim_{h\rightarrow0}\frac{f(1+h)}{h}=1$
so, $\displaystyle f(1)=0$ & $\displaystyle f'(1)=1$

2. ## Re: Differentiable function problem

Originally Posted by orir
$\displaystyle f$ is a differentiable function at $\displaystyle x=1$.prove that if $\displaystyle \lim_{h\rightarrow0}\frac{f(1+h)}{h}=1$
so, $\displaystyle f(1)=0$ & $\displaystyle f'(1)=1$

If $\displaystyle f(1)\ne 0$ is it possible for $\displaystyle {\lim _{h \to 0}}\frac{{f(1)}}{h}$ to exist? Why or why not?

3. ## Re: Differentiable function problem

i actually don't know what to answear to that.. :S

4. ## Re: Differentiable function problem

Originally Posted by orir
i actually don't know what to answear to that.. :S

Why don't you know that? It happens to be the key to this whole question.

What is $\displaystyle {\lim _{h \to 0}}~\frac{3}{h}=~?$ Can you work that out?

5. ## Re: Differentiable function problem

this lim doesn't exist... so? how this helps me?

6. ## Re: Differentiable function problem

Originally Posted by orir
this lim doesn't exist... so? how this helps me?
You are given that both
$\displaystyle {\lim _{h \to 0}}\frac{{f(1 + h)}}{h}\quad \& \quad {\lim _{h \to 0}}\frac{{f(1 + h) - f(1)}}{h}$ EXIST.

Now doesn't that imply that $\displaystyle {\lim _{h \to 0}}\frac{{f(1)}}{h}$ must also exists?

If so, what does that tell you about $\displaystyle f(1)~?$ And why?

7. ## Re: Differentiable function problem

i guess it tells me that $\displaystyle f(1)=0$...

8. ## Re: Differentiable function problem

Originally Posted by orir
i guess it tells me that $\displaystyle f(1)=0$...

Correct! So what is $\displaystyle f'(1)~?$

9. ## Re: Differentiable function problem

1! thank you...