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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?

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    $\begingroup$ Did you try rewriting as $a^b=e^{b\ln a}$ where $a=\frac{n+1}{n-2}$ and $b=\sqrt n$? Then try using L'Hospital's rule? $\endgroup$ – Gregory Grant Jan 29 '16 at 22:47
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$$\lim_{n\to\infty}\left({\frac{n+1}{n-2}}\right)^\sqrt n=\lim_{n\to\infty}\left(1+{\frac{3}{n-2}}\right)^{\frac{n-2}{3}\frac{3\sqrt n}{n-2}}=e^{0}=1$$ because $$\lim_{n\to\infty}\left(1+{\frac{3}{n-2}}\right)^{\frac{n-2}{3}}=e$$ $$\lim_{n\to\infty}\frac{3\sqrt n}{n-2}=0$$

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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.

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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 $\frac{\sqrt n}{n-2}\to 0$.

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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.

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  • $\begingroup$ would the downvoter comment pls? $\endgroup$ – Alex Jan 29 '16 at 22:52
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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

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

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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 $e^0=1$.

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