Answer
$e^{ab}$
Work Step by Step
$ \lim\limits_{x \to \infty} (1+ \frac{a}{x})^{bx}$
Let $y = (1+ \frac{a}{x})^{bx}$
$ \ln(y) = \ln(1+ \frac{a}{x})^{bx}$
$ \lim\limits_{x \to \infty} (1+ \frac{a}{x})^{bx} = \lim\limits_{x \to \infty} (e^{\ln(1+ \frac{a}{x})^{bx}}) = \lim\limits_{x \to \infty} (e^{bx\ln(1+ \frac{a}{x})})$
Next:
let $c = {bx\ln(1+ \frac{a}{x})}$
$\lim\limits_{x \to \infty} c =\frac{ {ln(1+ \frac{a}{x})}}{\frac{1}{bx}}$
which evaluates to $\frac{0}{0}$ so L'Hospital's Rule may be used giving :
$\lim\limits_{x \to \infty} \frac{ {\frac{1}{1+ \frac{a}{x}}(\frac{-a}{x^{2}})}}{\frac{-1}{bx^{2}}}$
by cancelling out the $x^{2}$ on the topmost fraction and the bottom most fraction we get:
$\lim\limits_{x \to \infty}\frac{ab}{1+\frac{a}{x}} = ab$
Plugging this into $e^{\ln(1+ \frac{a}{x})^{bx}}$ we get the final answer of:
$e^{ab}$
