Answer
$\approx 1$
Work Step by Step
We have to evaluate the limit:
$\lim\limits_{h \to 0} f(h)$, where $f(h)=\dfrac{\ln (1+h)}{h}$
We compute the value of the function $f$ for values of $h$ close to 0:
$f(0.01)\approx 0.99503309$
$f(0.001)\approx 0.99950033$
$f(0.0001)\approx 0.99995$
$f(-0.01)\approx 1.0050336$
$f(-0.001)\approx 1.0005003$
$f(-0.0001)\approx 1.00005$
We got $\lim\limits_{h \to 0} f(h)\approx 1$
