Find $\lim_{x\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}$

Question

Find $$\lim_{n\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}.$$

My attempt:

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

Now we should do something to change the power to $\frac{n-1}{3}$ because:

$\lim_{x\to \infty}(1+\frac{1}{x})^x=e$

But I cannot get the answer(the answer is $e^2$). Please give small hints not full answers.

Here is a picture from my answer:

Note that in persian $2=۲$ and $3=۳$.

edit:The answer is mistaked and take $n+1$ instad of $n+1$.

$\ln$, the natural log. $\ln(x)=\log_e(x)$ – Jul 25 '16 at 05:34 — , Jul 25 '16 at 05:34
@Zachary Selk How should I use L'Hopital's rule? – Taha Akbari Jul 25 '16 at 05:37 — Taha Akbari, Jul 25 '16 at 05:37

JasonM · Accepted Answer · 2016-07-25T05:41:48.450

3

$$\begin{align}\left(\frac{n+2}{n-1}\right)^{2n+3}&= \left(1+\frac{3}{n-1}\right)^{2n+3}\\ &= \left(1+\frac{3}{n-1}\right)^{2(n-1)+5}\\ &= \left(\left(1+\frac{3}{n-1}\right)^{n-1}\right)^2\left(1+\frac{3}{n-1}\right)^5\end{align}$$

If you like, you can replace $n-1$ with $n$ as $n\rightarrow \infty$

edited Jul 25 '16 at 05:41

answered Jul 25 '16 at 05:38

JasonM

3,151

Nice answer is there anywhere to learn how to find the limits that are increasing to $\infty$? – Taha Akbari Jul 25 '16 at 05:41
It is clear thats why I gave upvote. – Taha Akbari Jul 25 '16 at 05:43
@TahaAkbari I'm not sure what you mean by "find limits that are increasing to $\infty$" – JasonM Jul 25 '16 at 05:44
I mean $\lim_{n\to \infty}$. – Taha Akbari Jul 25 '16 at 05:45
@TahaAkbari Yes of course. $\left(1+\frac{a}{n}\right)^{n}\rightarrow e^a$ as $n\rightarrow \infty$, and $\frac{a}{n} \rightarrow 0$ as $n\rightarrow \infty$. – JasonM Jul 25 '16 at 05:50
but $\lim_{n \to \infty}\left(\left(1+\frac{3}{n-1}\right)^{n-1}\right)^2=(e^3)^2=e^6$, this is right? – Jul 25 '16 at 05:51
@juliogodoy Yes, $e^6$ is the correct answer, and not $e^2$ as the OP originally said. – JasonM Jul 25 '16 at 05:56
I put a persian language answer. Note that 2=۲ ans 3=۳. – Taha Akbari Jul 25 '16 at 06:00
I think the answer is mistaked.I take $n+1$ instad of $n-1$ – Taha Akbari Jul 25 '16 at 06:04
@TahaAkbari If it were $n+1$, then the answer would be $e^2$. Careful with that – JasonM Jul 25 '16 at 06:05
Yes you are right the answer was wrong. – Taha Akbari Jul 25 '16 at 06:06

score 2 · Answer 2 · answered Jul 25 '16 at 06:30

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\lim_{n \to \infty}\pars{n + 2 \over n - 1}^{2n + 3}} & = \lim_{n \to \infty}\,\bracks{% \pars{1 + 2/n \over 1 - 1/n}^{2n}\pars{1 + 2/n \over 1 - 1/n}^{3}} \\[5mm] & = \lim_{n \to \infty}\,\braces{\bracks{\pars{1 + {2 \over n}}^{n}}^{2} \bracks{\pars{1 - {1 \over n}}^{n}}^{-2}} \\[5mm] & = \bracks{\lim_{n \to \infty}\pars{1 + {2 \over n}}^{n}}^{2} \bracks{\lim_{n \to \infty}\pars{1 - {1 \over n}}^{n}}^{-2} = \pars{\expo{2}}^{2}\pars{\expo{-1}}^{-2} \\[5mm] & = \color{#f00}{\expo{6}} \approx 403.4288 \end{align}

score 1 · Answer 3 · answered Jul 25 '16 at 05:48

Hint:

$$ L = \lim_{x \ \rightarrow \ c} f(x)^{g(x)} \implies \log \left ( L \right ) = \log \left ( \lim_{x \ \rightarrow \ c} f(x)^{g(x)} \right ) $$ Then by the continuity of $\log(x)$ and using $\log_{b}a^x = x\log_{b}a$: $$ \log \left ( L \right ) = \lim_{x \ \rightarrow \ c} \left [ g(x) \log \left ( f(x) \right ) \right ] $$

score 0 · Answer 4 · answered Jul 25 '16 at 06:15

$$\lim _{ n\to \infty } \left( \frac { n+2 }{ n-1 } \right) ^{ 2n+3 }=\lim _{ n\to \infty } \left( 1+\frac { 3 }{ n-1 } \right) ^{ 2n+3 }=\\ =\lim _{ n\to \infty }{ \left[ \left( 1+\frac { 1 }{ \frac { n-1 }{ 3 } } \right) ^{ \frac { n-1 }{ 3 } } \right] } ^{ \frac { 3 }{ n-1 } \left( 2n+3 \right) }=\lim _{ n\rightarrow \infty }{ { e }^{ \frac { 6n+9 }{ n-1 } } } ={ e }^{ 6 }$$

score 0 · Answer 5 · answered Jul 25 '16 at 06:59

Just another possible way to do it.

Considering $$a_n=\left(\frac{n+2}{n-1}\right)^{2n+3}$$ take logarithms $$\log(a_n)=(2n+3)\log\left(\frac{n+2}{n-1}\right)=(2n+3)\log\left(1+\frac 3{n-1}\right)$$ Now, using Taylor series for small $x$ $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$ replace $x$ by $\frac 3{n-1}$ followed by long division to get $$\log\left(1+\frac 3{n-1}\right)=\frac{3}{n}-\frac{3}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ which makes $$\log(a_n)=6+\frac{6}{n}+O\left(\frac{1}{n^2}\right)$$ and using $a_n=e^{\log(a_n)}$ and Taylor again$$a_n=e^6+\frac{6 e^6}{n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.

Find $\lim_{x\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}$

5 Answers5