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:


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!


There is no one generic technique which applies to all cases, but it's often useful to recognize known patterns and try to reduce the problem to an equivalent, but easier one, using those patterns.

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:

$$\lim_{x \to \infty} \left(1+\frac{a}{x}\right)^x = \lim_{x \to \infty} \left(\left(1+\frac{1}{\frac{x}{a}}\right)^{\frac{x}{a}}\right)^a = \left(\lim_{\frac{x}{a} \to \infty} \left(1+\frac{1}{\frac{x}{a}}\right)^{\frac{x}{a}}\right)^a = e^a$$

The above still needs a few justifications for the intermediate steps in order to become a complete proof, but it's almost there, and required no tedious calculations along the way.


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