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I'm having difficulties with finding this limit: $$\lim_{n\to\infty}\left(1-\frac{2}{5n+5}\right)^{3n}$$

According to my notes from lecture I need to make it to this form: $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e$$

I have no bloody idea how to solve this, but I've solved similar exercises like: $$\lim_{n\to\infty}\left(1 + \frac{2}{n}\right)^n = e^2$$ and so on.

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Hint. Note that $$\left(1-\frac{2}{5n+5}\right)^{3n}=\left(\left(1-\frac{2}{5n+5}\right)^{-\frac{5n+5}{2}}\right)^{-\frac{6n}{5n+5}}.$$

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  • $\begingroup$ Oh, it get it now. Now I see what was done there. Thank you. $\endgroup$ Nov 11 '16 at 9:03
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write $$(1- \frac{2}{5n+5})^{3n}=e^{\log_e(1- \frac{2}{5n+5})^{3n} }=e^{3n\log_e(1- \frac{2}{5n+5}) }.$$

When $n$ goes to infinity $\log(1- \frac{2}{5n+5})$ is asymptotic to $\frac{-2}{5n+5}$. So you get $$e^{3n \log_e(1- \frac{2}{5n+5})}\sim e^{\frac{-6n}{5n+5}} \rightarrow e^{\frac{-6}{5}}$$ when $n\rightarrow + \infty$.

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$$\left(1-\dfrac2{5n+5}\right)^{3n}=\left(\left(1-\dfrac2{5n+5}\right)^{-(5n+5)/2}\right)^{-6n/5n+5}$$

Now $$\lim_{n\to\infty}\dfrac{-6n}{5n+5}=\dfrac{-6}{5+\lim_{n\to\infty}\dfrac5n}=?$$

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$$\lim_{n\to\infty}(1-\frac{2}{5n+5})^{3n}=\lim_{n\to\infty}\left((1-\frac{\frac{2}{5}}{n+1})^{n+1}\right)^3\lim_{n\to\infty}\left((1-\frac{\frac{2}{5}}{n+1})^{-1}\right)^3=\lim_{n\to\infty}\left((1-\frac{\frac{2}{5}}{n+1})^{n+1}\right)^3(1)=e^{-\frac{6}{5}}$$

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