Answer
$${e^{ab}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{a}{x}} \right)^{bx}},{\text{ }}a,b{\text{ > 0}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{a}{x}} \right)^{bx}} = {\left( {1 + \frac{a}{{ + \infty }}} \right)^\infty } = {1^\infty } \cr
& {\text{This limit has the form }}{1^\infty }{\text{ }} \cr
& {\left( {1 + \frac{a}{x}} \right)^{bx}} = {e^{bx\ln \left( {1 + \frac{a}{x}} \right)}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{a}{x}} \right)^{bx}} = \mathop {\lim }\limits_{x \to + \infty } {e^{bx\ln \left( {1 + \frac{a}{x}} \right)}} = {e^{\mathop {\lim }\limits_{x \to + \infty } bx\ln \left( {1 + \frac{a}{x}} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to + \infty } bx\ln \left( {1 + \frac{a}{x}} \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{\ln \left( {1 + \frac{a}{x}} \right)}}{{\frac{1}{{bx}}}} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to + \infty } \frac{{\ln \left( {1 + \frac{a}{x}} \right)}}{{\frac{1}{{bx}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{ - \frac{a}{{{x^2}}}}}{{1 + \frac{a}{x}}}}}{{ - \frac{1}{{b{x^2}}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{ - \frac{a}{{{x^2}}}}}{{\frac{{x + a}}{x}}}}}{{ - \frac{1}{{b{x^2}}}}} \cr
& L = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{{ - a}}{{x\left( {x + a} \right)}}}}{{ - \frac{1}{{b{x^2}}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{ab{x^2}}}{{x\left( {x + a} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{abx}}{{x + a}} \cr
& \mathop {\lim }\limits_{x \to + \infty } \frac{{abx}}{{x + a}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{ab}}{{1 + \frac{a}{x}}} \cr
& \mathop {\lim }\limits_{x \to + \infty } \frac{{ab}}{{1 + \frac{a}{x}}} = \frac{{ab}}{{1 + \frac{a}{\infty }}} = ab \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to + \infty } {\left( {1 + \frac{a}{x}} \right)^{bx}} = \mathop {\lim }\limits_{x \to + \infty } {e^{bx\ln \left( {1 + \frac{a}{x}} \right)}} = {e^{ab}} \cr} $$
