TITLE: Finding the limit $\lim_{n\to \infty} \left({\frac{n+1}{n-2}}\right)^\sqrt n$
QUESTION [6 upvotes]: I have to find:
$$\lim_{n\to \infty} \left({\frac{n+1}{n-2}}\right)^\sqrt n$$
But, to be honest, I haven't got a faintest idea how to even begin. Is there a way to evaluate this radical exponent?

REPLY [3 votes]: $$\lim_{n\to\infty}\left({\frac{n+1}{n-2}}\right)^\sqrt n=\lim_{n\to\infty}\left(1+{\frac{3}{n-2}}\right)^{\frac{n-2}{3}\frac{3\sqrt n}{n-2}}=e^{0}=1$$
because
$$\lim_{n\to\infty}\left(1+{\frac{3}{n-2}}\right)^{\frac{n-2}{3}}=e$$
$$\lim_{n\to\infty}\frac{3\sqrt n}{n-2}=0$$

REPLY [2 votes]: Note that we have
$$\left(\frac{n+1}{n-2}\right)^{\sqrt n}=\left(\frac{\left(1+\frac1n\right)^n}{\left(1-\frac2n\right)^n}\right)^{n^{-1/2}}$$
In THIS ANSWER and THIS ONE, I showed using only the limit definition of the exponential function that $\left(1+\frac xn\right)^n$ is  monotonically increasing for $x>-n$.  Therefore, we have
$$2\le \left(1+\frac1n\right)^n<e \tag 1$$
for $n\ge 1$ and for $n\ge 4$
$$e^{-2}> \left(1-\frac2n\right)^n\ge \frac1{16} \tag 2$$
Putting $(1)$ and $(2)$ together, we find
$$(2e^2)^{n^{-1/2}}\le \left(\frac{n+1}{n-2}\right)^{\sqrt n}\le (16e)^{n^{-1/2}}$$
whereupon applying the squeeze theorem yields
$$\lim_{n\to \infty}\left(\frac{n+1}{n-2}\right)^{\sqrt n}=1$$