# Thread: Show definition of derivative

1. ## Show definition of derivative

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

2. Originally Posted by wopashui
Show that $\displaystyle lim_x->0$$\displaystyle ln (1+x)/x = 1$from the definition of derivative.
Hello
$\displaystyle \lim_{x\to 0} \frac{ln(1+x)}{x} = \lim_{x\to 0} \frac{ln(1+x)-0}{x-0}$
Let $\displaystyle f(x)=ln(1+x)$
Clearly $\displaystyle f(0)=0$
So our limit can be written as:
$\displaystyle \lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)$

Got it?

3. Originally Posted by General
Hello
$\displaystyle \lim_{x\to 0} \frac{ln(1+x)}{x} = \lim_{x\to 0} \frac{ln(1+x)-0}{x-0}$
Let $\displaystyle f(x)=ln(1+x)$
Clearly $\displaystyle f(0)=0$
So our limit can be written as:
$\displaystyle \lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)$

Got it?

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

4. Originally Posted by wopashui
so that produce f(x)/x, why is it equal to 1?
$\displaystyle \frac{ln(1+x)}{x}=\frac{ln(1+x)-0}{x-0}$
Do you have problem in this one?
Substracting zero is allowed everywhere.
and I replaced $\displaystyle 0$ by $\displaystyle f(0)$
since $\displaystyle f(0)=ln(1+0)=ln(1)=0$

$\displaystyle \frac{ln(1+x)}{x}=\frac{ln(1+x)-0}{x-0}$
and I replaced $\displaystyle 0$ by $\displaystyle f(0)$
since $\displaystyle f(0)=ln(1+0)=ln(1)=0$