# Differentiable function problem

• Apr 20th 2013, 10:16 AM
orir
Differentiable function problem
$f$ is a differentiable function at $x=1$.prove that if $\lim_{h\rightarrow0}\frac{f(1+h)}{h}=1$
so, $f(1)=0$ & $f'(1)=1$
• Apr 20th 2013, 10:26 AM
Plato
Re: Differentiable function problem
Quote:

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

If $f(1)\ne 0$ is it possible for ${\lim _{h \to 0}}\frac{{f(1)}}{h}$ to exist? Why or why not?
• Apr 20th 2013, 10:49 AM
orir
Re: Differentiable function problem
i actually don't know what to answear to that.. :S
• Apr 20th 2013, 10:53 AM
Plato
Re: Differentiable function problem
Quote:

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 ${\lim _{h \to 0}}~\frac{3}{h}=~?$ Can you work that out?
• Apr 20th 2013, 12:03 PM
orir
Re: Differentiable function problem
this lim doesn't exist... so? how this helps me?
• Apr 20th 2013, 12:12 PM
Plato
Re: Differentiable function problem
Quote:

Originally Posted by orir
this lim doesn't exist... so? how this helps me?

You are given that both
${\lim _{h \to 0}}\frac{{f(1 + h)}}{h}\quad \& \quad {\lim _{h \to 0}}\frac{{f(1 + h) - f(1)}}{h}$ EXIST.

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

If so, what does that tell you about $f(1)~?$ And why?
• Apr 20th 2013, 12:40 PM
orir
Re: Differentiable function problem
i guess it tells me that $f(1)=0$...
• Apr 20th 2013, 12:44 PM
Plato
Re: Differentiable function problem
Quote:

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

Correct! So what is $f'(1)~?$
• Apr 20th 2013, 12:51 PM
orir
Re: Differentiable function problem
1! thank you...