TITLE: Limit with x approaching infinity
QUESTION [5 upvotes]: The problem says:

If $$\lim_{x\to +\infty} \left\lbrack\frac{ax+1}{ax-1}\right\rbrack^x=9$$, determine $a$.

It appears to be a case of $\left\lbrack\frac{\infty}{\infty}\right\rbrack^\infty$. How can I solve such a case?

REPLY [6 votes]: This is definetely a very interesting question. Here is what I think is a good way to think about it. First rearrange the expression (omitting limits for clarity):
$$\left( \frac{a x + 1 - 2 + 2 }{a x -1} \right)^x = \left( 1 + \frac{ 2 }{a x -1} \right)^x $$
Notice that, in the limit of $x \rightarrow ∞ $, the denominator reduces to $a x$. We rewrite the equation as:
$$\lim_{x\to \infty} \left( 1 + \frac{ 2 }{a x -1} \right)^x = 9 $$
or by noting what we have said about the denominator:
$$\lim_{x\to \infty} \left( 1 + \frac{ 2 }{a x} \right)^x = 9 $$
The LHS is exactly the definition of $e^{2/a}$, hence you can say that:
$$e^{2/a} = 9$$
from which you can easily find that:
$$a = \frac{2}{\ln{9}} = \frac{1}{\ln{(3)}}$$
I hope this helps you!