Answer
$\lim\limits_{x \to 0}\frac{ln(1+x)}{x}=1$
Work Step by Step
Consider $f(x)=lnx$
and $f'(x)=\frac{1}{x}$
Also,
$f'(1)=1$
From the definition of derivative as a limit, we get
$f'(1) =\lim\limits_{h \to 0}\frac{f(1+h)-f(1)}{h}$
Replacing $ h$ by $x$.
$f'(1) =\lim\limits_{x \to 0}\frac{f(1+x)-f(1)}{x}$
$=\lim\limits_{x \to 0}\frac{ln(1+x)-ln(1)}{x}$, where $f(x)=lnx$
Thus,
$f'(1) =\lim\limits_{x \to 0}\frac{ln(1+x)}{x}$
Since,
$f'(1)=1$
Hence,$\lim\limits_{x \to 0}\frac{ln(1+x)}{x}=1$
