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I need help to calculate this limit:

$$\lim_{n \to \infty}\frac{n^{3}}{(3+\frac{1}{n})^{n}}$$

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  • $\begingroup$ Is the limit $1$ $\endgroup$ – Archis Welankar Dec 18 '15 at 10:42
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$$0 \le \frac{n^3}{(3+\frac{1}{n})^n} \le \frac{n^3}{3^n} $$ $$\lim_{n\to \infty}\frac{n^3}{3^n} = 0$$ So $$\lim_{n \to \infty} \frac{n^3}{(3+\frac{1}{n})^n} = 0$$

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And this is why $\lim_{n\to \infty}\frac{n^3}{3^n} = 0$. ${3^n}=(1+2)^n=1+n2+\frac{n(n-1)}{2}2^2+\frac{n(n-1)(n-2)}{3!}2^3+\frac{n(n-1)(n-2)(n-3)}{4!}2^4+...+2^n>\frac{n(n-1)(n-2)(n-3)}{4!}$

And so,

$0 \le \lim_{n\to \infty}\frac{n^3}{3^n} \le \lim_{n\to \infty}\frac{n^3}{\frac{n(n-1)(n-2)(n-3)}{4!}}= 0$

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