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

I need help to calculate this limit:

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

• Is the limit $1$ – Archis Welankar Dec 18 '15 at 10:42

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