Finding the limit $\lim_{n\to \infty} \left({\frac{n+1}{n-2}}\right)^\sqrt n$ 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?
 A: 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.
A: 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 $\frac{\sqrt n}{n-2}\to 0$.
A: 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.
A: $$\lim_{n\to\infty}\left({\frac{n+1}{n-2}}\right)^\sqrt n=\lim_{n\to\infty}\left(1+{\frac{3}{n-2}}\right)^{\frac{n-2}{3}\frac{3\sqrt n}{n-2}}=e^{0}=1$$
because
$$\lim_{n\to\infty}\left(1+{\frac{3}{n-2}}\right)^{\frac{n-2}{3}}=e$$
$$\lim_{n\to\infty}\frac{3\sqrt n}{n-2}=0$$
A: 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
$$\lim_{n\to \infty}\left(\frac{n+1}{n-2}\right)^{\sqrt n}=1$$
A: A simple way around is to compute the limit of the logarithm and to replace the sequence by a function that has the same values on the integers:
$$
\lim_{x\to\infty}\sqrt{x}\log\dfrac{x+1}{x-2}
$$
Substitute $x=1/t$, so you have
$$
\lim_{t\to0^+}\sqrt{t}\log\dfrac{1+t}{1-2t}=
\lim_{t\to0^+}\frac{\log(1+t)-\log(1-2t)}{\sqrt{t}}=
\lim_{t\to0^+}\frac{\log(1+t)-\log(1-2t)}{t}\sqrt{t}
$$
Now
$$
\lim_{t\to0^+}\frac{\log(1+t)-\log(1-2t)}{t}
$$
is the derivative at $0$ of $f(t)=\log(1+t)-\log(1-2t)$:
$$
f'(t)=\frac{2}{1+t}+\frac{1}{1-2t},
\qquad
f'(0)=3
$$
So the limit is $0$ and the one you're looking for is $e^0=1$.
