Answer
$$\lim_{x\to\infty}\Bigg(1+\frac{a}{x}\Bigg)^{bx}=e^{ab}$$
Work Step by Step
$$A=\lim_{x\to\infty}\Bigg(1+\frac{a}{x}\Bigg)^{bx}$$
We take the natural logarithm of both sides:
$$\ln A=\ln\Bigg[\lim_{x\to\infty}\Bigg(1+\frac{a}{x}\Bigg)^{bx}\Bigg]$$
$$\ln A=\lim_{x\to\infty}\Bigg[\ln\Bigg(1+\frac{a}{x}\Bigg)^{bx}\Bigg]$$
$$\ln A=\lim_{x\to\infty}bx\ln \Bigg(1+\frac{a}{x}\Bigg)$$
$$\ln A=\lim_{x\to\infty}\frac{bx\ln\Bigg(1+\frac{a}{x}\Bigg)}{1}.$$
Divide both numerator and denominatory by $x$
$$\ln A=\lim_{x\to\infty}\frac{b\ln(1+\frac{a}{x})}{\frac{1}{x}}.$$
Now we take $\frac{1}{x}=u$. So as $x\to\infty$, we have $u\to0$.
$$\ln A=\lim_{u\to0}\frac{b\ln(1+au)}{u}$$
Because $\lim_{u\to0}[b\ln(1+au)]=b\ln(1+a\times0)=b\ln1=b\times0=0$ and $\lim_{u\to0}u=0$,
we have an indeterminate form of $\frac{0}{0}$. Following L'Hospital's Rule, we would have
$$\ln A=\lim_{u\to0}\frac{b\times\frac{1}{1+au}\times(1+au)'}{1}$$
$$\ln A=\lim_{u\to0}\frac{b}{1+au}\times a$$
$$\ln A=\lim_{u\to0}\frac{ab}{1+au}$$
$$\ln A=\frac{ab}{1+a\times0}$$
$$\ln A=\frac{ab}{1}=ab$$
$$A=e^{ab}$$
