Answer
$\approx 7.39=e^2$
Work Step by Step
We have to evaluate the limit:
$\lim\limits_{h \to 0} f(h)$, where $f(h)=(1+2h)^{1/h}$
We compute the value of the function $f$ for values of $h$ close to 0:
$f(0.01)\approx 7.2446461$
$f(0.001)\approx 7.3743124$
$f(0.0001)\approx 7.3875786$
$f(-0.01)\approx 7.5403661$
$f(-0.001)\approx 7.4038688$
$f(-0.0001)\approx 7.3905343$
We got $\lim\limits_{h \to 0} f(h)\approx 7.39=e^2$
