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I'm struggling with this assignment for a couple of hours, and i dont seem to find any clues :(

I have to find the limit of this exponential function: $$\lim_{n \to \infty} (1+\frac{1}{n+2})^{2n+3} $$

I've already tried to rewrite the expression as: $$\lim_{n \to \infty} ((1+\frac{1}{n+2})^{n+2})^{\frac{2n+3}{n+2}}$$ $$\lim_{n \to \infty} \frac{2n+3}{n+2} = 2 $$ $$\lim_{n \to \infty} (1+\frac{1}{n+2})^{2n+4}$$ $$k=n+2$$ $$\lim_{n \to \infty} (1+\frac{1}{k})^{2k})$$

But it didn't help actually. I know i have to rewrite the limit to the form of: $$\lim_{n \to \infty} (1+\frac{x}{n})^{n}= e^x$$ Unfortunately i'm out of ideas now. I will really appreciate any tips from you guys. I totally stuck with this one.

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You have solved the problem yourself. Just check this.

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

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  • $\begingroup$ Where is the fault in my reasoning? We have: $2n+3=(n+2)+(n+1)$. So $$\lim_{n \to \infty} (1+\frac{1}{n+2})^{2n+3}=\lim_{n \to +\infty}\left( \underbrace{(1+\frac{1}{n+2})^{n+2}}_\text{$\to e$ as $n \to +\infty$}\cdot (1+\frac{1}{n+2})^{n+1}\right)$$ i.e. if the limit equals $e^2$, then $\lim_{n \to +\infty} (1+\frac{1}{n+2})^{n+1}$ should go to $e$, but it clearly does not. $\endgroup$
    – derthomas
    Nov 21 '15 at 12:52
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    $\begingroup$ @derthomas It truly goes to $e$. $\large\frac{\lim_{n\to \infty}(1+\frac{1}{n+2})^{(n+2)}}{\lim_{n\to \infty}(1+\frac{1}{n+2})}=\frac{e}{1}$ $\endgroup$ Nov 21 '15 at 12:59
  • $\begingroup$ Oh yes, I see. The source of my faulty reasoning was in thinking that $$\left(\left(\lim_{n\to+\infty}f(x)=e\right) \wedge \left(\lim_{n\to+\infty}g(x)=e\right)\right) \implies f(x)=g(x)$$ which is complete non sense $\endgroup$
    – derthomas
    Nov 21 '15 at 13:18
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    $\begingroup$ @derthomas You're welcome. $\endgroup$ Nov 21 '15 at 13:19

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