6
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
3
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
6
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ would the downvoter comment pls? $\endgroup$ – Alex Jan 29 '16 at 22:52
2
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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 $e^0=1$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.