Answer
$e^{ab}$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to \infty}(1+\dfrac{a}{x})^{bx}$
Let us consider that $y=(1+\dfrac{a}{x})^{bx} \implies \ln y=bx \ln (1+\dfrac{a}{x})$
But $\lim\limits_{x \to \infty} \dfrac{b \ln (1+\dfrac{a}{x})}{1/x}=\dfrac{0}{0}$
We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{0}{0}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to \infty} \ln y=\lim\limits_{x \to \infty} \dfrac{b}{(1+a/x)} \times \dfrac{(-a/x^2)}{(-1/x^2)}\\= ab$
So, $\lim\limits_{x \to \infty} \ln y=ab \implies \lim\limits_{x \to \infty} y=e^{ab}$
