Show definition of derivative

**wopashui** · January 28th 2010, 03:58 PM

Show that $lim_{x-->0}$ $ln (1+x)/x$ = 1 from the definition of derivative.

**General** · January 28th 2010, 04:03 PM

Originally Posted by wopashui

Show that $lim_x->0$ $ln (1+x)/x = 1$ from the definition of derivative.

Hello

$\lim_{x\to 0} \frac{ln(1+x)}{x} = \lim_{x\to 0} \frac{ln(1+x)-0}{x-0}$
Let $f(x)=ln(1+x)$
Clearly $f(0)=0$
So our limit can be written as:
$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)$

Got it?

**wopashui** · January 28th 2010, 05:52 PM

Originally Posted by General

Hello

$\lim_{x\to 0} \frac{ln(1+x)}{x} = \lim_{x\to 0} \frac{ln(1+x)-0}{x-0}$
Let $f(x)=ln(1+x)$
Clearly $f(0)=0$
So our limit can be written as:
$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)$

Got it?

so that produce f(x)/x, why is it equal to 1?

**General** · January 28th 2010, 05:57 PM

Originally Posted by wopashui

so that produce f(x)/x, why is it equal to 1?

I did not follow you!
$\frac{ln(1+x)}{x}=\frac{ln(1+x)-0}{x-0}$
Do you have problem in this one?
Substracting zero is allowed everywhere.
and I replaced $0$ by $f(0)$
since $f(0)=ln(1+0)=ln(1)=0$

Please, explain more.

**wopashui** · January 30th 2010, 11:22 AM

Originally Posted by General

I did not follow you!
$\frac{ln(1+x)}{x}=\frac{ln(1+x)-0}{x-0}$
Do you have problem in this one?
Substracting zero is allowed everywhere.
and I replaced $0$ by $f(0)$
since $f(0)=ln(1+0)=ln(1)=0$

Please, explain more.

I meant, we are proving the limit is equal to 1 not 0, how does that make sense?

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