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I am having a hard time finding the next step on this.

Can someone tell me what is the next step and explain to me in non-mathematical terms what I am supposed to do when I reach this examples at this place ?

\begin{align*} \lim_{n\to\infty}\left(\dfrac{n+3}{n+1}\right)^{2n} &=\lim_{n\to\infty}\left(\dfrac{n+1+2}{n+1}\right)^{2n}\\ &=\lim_{n\to\infty}\left(1+\dfrac{2}{n+1}\right)^{2n}\\ &=\lim_{n\to\infty}\left(1+\dfrac{2}{n+1}\right)^{?} \end{align*}

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  • $\begingroup$ You must use the known limit $\lim_{n\to+\infty}(1+1/n)^n=e$. $\endgroup$
    – User3773
    Feb 19 '15 at 15:10
  • $\begingroup$ i know that, but you can't use that limit because the bottom part is n+1 and the top one is 2n. What i asked is how do i get to make the bottom part or the top one same as the other ? $\endgroup$
    – John
    Feb 19 '15 at 15:11
  • $\begingroup$ use the fact that $\lim_{n \to \infty}(1+x/n)^n = e^x$ $\endgroup$
    – abel
    Feb 19 '15 at 15:15
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$$\lim_{n\to\infty}\left(\dfrac{n+3}{n+1}\right)^{2n} = \lim_{n\to\infty}\left ( \frac{n\cdot \left (1+\frac{3}{n} \right )}{n\cdot\left (1+\frac{1}{n} \right ) } \right )^{2n} = \lim_{n\to\infty}\left ( \frac{\left (1+\frac{3}{n} \right )}{\left (1+\frac{1}{n} \right ) } \right )^{2n} = \lim_{n\to\infty}\frac{\left (\left ( 1+\frac{3}{n} \right )^{n}\right)^2}{\left (\left ( 1+\frac{1}{n} \right )^{n} \right )^{2}} =$$$$ = \frac{\left (e^{3} \right )^{2}}{\left (e^{1} \right )^{2}} = \frac{e^6}{e^2} = e^{6-2} = e^4.$$

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  • $\begingroup$ wow this is so much easier for me to be understood $\endgroup$
    – John
    Feb 19 '15 at 15:40
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HINT:

  • $e=(1+\frac1x)^x$

  • $(1+\frac{2}{n+1})^{2n}=(1+\frac{2}{n+1})^{\frac{n+1}{2}\cdot4-2}=\frac{\left((1+\frac{2}{n+1})^{\frac{n+1}{2}}\right)^4}{(1+\frac{2}{n+1})^2}$

OK, apparently some more hints are required, so here is another one:

$$\lim\limits_{x\to\infty}\frac{f(x)^p}{g(x)^q}=\frac{\left(\lim\limits_{x\to\infty}f(x)\right)^p}{\left(\lim\limits_{x\to\infty}g(x)\right)^q}$$

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  • $\begingroup$ I hate the fact that logically i can not understand why do you do this step, and what makes you think on such way that leads you to doing it like this. Can you please share the rest of the equation with me, i can not figure out what to do next $\endgroup$
    – John
    Feb 19 '15 at 15:33
  • $\begingroup$ @John: I added another hint. Please let me know if it helps (or if it doesn't)... $\endgroup$ Feb 19 '15 at 15:41
  • $\begingroup$ @John $\cdots = \frac { \left( \left(1+\frac{1}{\frac{n+1}{2}}\right)^{\frac{n+1}{2}} \right)^4 }{\left(1+\frac{2}{n+1}\right)^2}$ $\endgroup$
    – user26486
    Feb 19 '15 at 15:46
  • $\begingroup$ @user314: What's the point of copying a part of the answer into a comment below it? $\endgroup$ Feb 19 '15 at 15:52
  • $\begingroup$ @barakmanos I re-wrote the numerator of the expression to make $\left( 1 + \frac{1}{ \text {something}} \right) ^ { \text {something}}$ more apparent. My expression is slightly different from yours. $\endgroup$
    – user26486
    Feb 19 '15 at 15:55
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Your strategy is very good; you can make the substitution $m=n+1$ and your last limit becomes $$ \lim_{m\to\infty}\left(\left(1+\frac{2}{m}\right)^{\!m-1}\right)^{\!2}= \lim_{m\to\infty} \left(\left(1+\frac{2}{m}\right)^{\!m}\right)^{\!2} \left(1+\frac{2}{m}\right)^{\!-2}=e^4 $$

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You can use the fact that the derivative of $e^x$ at $x = 0$ is $1,$ and the additive properties of $e^x$ to get the useful approximation $$(1+small)^{BIG} = e^{small\cdot BIG}+\cdots $$

you have $$\left(1+\frac2{n+1} \right)^{2n} = e^{4n/(n+1)}+\cdots = e^4+\cdots $$

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Not an efficient way, but you could also put the limit into a form where L'Hospital's Rule can be used.

\begin{align*} &\lim_{n\to\infty}\left(\dfrac{n+3}{n+1}\right)^{2n}\\ &=\lim_{n\to\infty}(\operatorname{exp}\left(2n\ln\left(\dfrac{n+3}{n+1}\right)\right))\\ &\implies \operatorname{exp}\left(\lim_{n\to\infty}\left(2n\ln\left(\dfrac{n+3}{n+1}\right)\right)\right)\\ &=\operatorname{exp}\left(2\lim_{n\to\infty}\dfrac{\ln\left(\frac{n+3}{n+1}\right)}{\frac{1}{n}}\right)\\ &=\operatorname{exp}\left(2\lim_{n\to\infty}\dfrac{\frac{2}{n^2+4n+3}}{\frac{1}{n^2}}\right) \text{by L'Hospital's Rule}\\ &=\operatorname{exp}\left(2\lim_{n\to\infty}\dfrac{2n^2}{n^2+4n+3}\right)\\ &=\operatorname{exp}(2\cdot 2)=e^4 \end{align*}

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