## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to 0}\frac{ln(1+x)}{x}=1$
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, $f'(1) =\lim\limits_{x \to 0}\frac{ln(1+x)}{x}$ $f'(1)=1$ Hence,$\lim\limits_{x \to 0}\frac{ln(1+x)}{x}=1$