# Show definition of derivative

• Jan 28th 2010, 04:58 PM
wopashui
Show definition of derivative
Show that $lim_{x-->0}$ $ln (1+x)/x$ = 1 from the definition of derivative.
• Jan 28th 2010, 05:03 PM
General
Quote:

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?
• Jan 28th 2010, 06:52 PM
wopashui
Quote:

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?
• Jan 28th 2010, 06:57 PM
General
Quote:

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 $f(0)=ln(1+0)=ln(1)=0$

• Jan 30th 2010, 12:22 PM
wopashui
Quote:

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 $f(0)=ln(1+0)=ln(1)=0$