# More Efficient Way of Evaluating a Limit?

I just finished a question on indeterminate powers and I'm fed up with how slow and inefficient my method of evaluating limits at discontinuities and infinities are.

The question was $\lim_\limits{x\rightarrow\infty}\left(1+\dfrac{a}{x}\right)^{x}$. After using a few limit laws, I was able to get to the correct answer. Which is:

$\lim_\limits{x\rightarrow\infty}\left(1+\dfrac{a}{x}\right)^{x}=e^a$

I won't show all of my work here, but, I was able to get the expression into the form,

$\lim_\limits{x\rightarrow\infty}[f(x)]^{g(x)}= e^{\lim_\limits{x\rightarrow \infty}\frac{\ln(1+\frac{a}{x})}{x^{-1}}}$

I then used L’Hôpital's rule to get rid of that nasty indeterminate exponent and was able to get the exponent to look like this:

$\lim_\limits{x\rightarrow\infty}\frac{ax}{(x+a)}$ $\leftarrow$That was easy :)

My problem is, I don't know an efficient way of evaluating the limit as it approaches an infinity (or some discontinuity). So evaluating the limit of this exponent was a real pain for me. For about 4 years, to solve limits I've made a mental table of values and subbed in a few $x$-values towards pos/neg infinity, keeping track of the results of the outputs, then guesstimating which $y$-value they approach. If I'm lucky, it's a nice number... but usually its an irrational number or a crazy fraction:(

If anyone knows a more efficient way of evaluating limits, or even, if you evaluate limits a different way; I would love to know. Thanks a million!

For the given example, it should reminisce of the well known: $$\lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x = e$$
Difference in the given problem, though, is that it has $\frac{a}{x}$ instead of $\frac{1}{x}$ inside the base. But $\frac{a}{x}$ behaves similarly to $\frac{1}{x}$ when $x \to \infty$ in the sense that both $\to 0$. So maybe there is a way to rewrite the limit in terms of $\frac{a}{x}$ in such a way as to reduce it to the well known case. After some guessing and trying, the following may pop to mind:
$$\lim_{x \to \infty} \left(1+\frac{a}{x}\right)^x = \lim_{x \to \infty} \left(\left(1+\frac{1}{\frac{x}{a}}\right)^{\frac{x}{a}}\right)^a = \left(\lim_{\frac{x}{a} \to \infty} \left(1+\frac{1}{\frac{x}{a}}\right)^{\frac{x}{a}}\right)^a = e^a$$