Finding the limit $\lim_{n\to \infty} \left({\frac{n+1}{n-2}}\right)^\sqrt n$

Hint

A way to solve indeterminations $1^\infty$ is:

If $\lim_n a_n=1$ and $\lim_n b_n=\infty$ then $$\lim_n a_n^{b_n}=e^{\lim_n (a_n-1)b_n}.$$ You can apply this to your case.

Here it is: $$ \left({\frac{n+1}{n-2}}\right)^\sqrt n= \left[\left({1+\frac{3}{n-2}}\right)^{(n-2)}\right]^{\frac{\sqrt n}{n-2}}\to(e^3)^0=1 $$

in fact $ \left({1+\frac{3}{n-2}}\right)^{(n-2)}\to e^3$ and $\frac{\sqrt n}{n-2}\to 0$.

Hint 1: $\frac{n+1}{n-2} = 1 + \frac{3}{n-2}$

Hint 2: $t = e^{\log t}$

Hint 3: Taylor series expansion around 1.