Finding a basic limit Determine the following limit
$$\lim_{n\to\infty} \cos{\left(\frac{n}{2^{n}}\right)}$$
I'm not really sure how to start here. We can write this as $$\cos{\left(\lim_{n\to\infty}\frac{n}{2^{n}}\right)}$$ then we must determine $$\lim_{n\to\infty}\frac{n}{2^{n}}$$ evidently this is zero however how would we actually determine this?
 A: The answer is $1$ because:


*

*The function $\cos x$ is continous on $\mathbb{R}$ and at $x = 0$.

*$\displaystyle \lim_{x\to 0} \cos x = \cos 0 = 1$.

*$\displaystyle \lim_{n \to \infty} \dfrac{n}{2^n} = 0$ ( by L'hospitale rule )
A: Do you know that $\lim\limits_{x\to a}f(g(x))=\lim\limits_{u\to b}f(u)$ where $b=\lim\limits_{x\to a}g(x)$ for sufficiently well-behaved $f,g$?

To prove $n/2^n\to0$ just recall that $2^n>n^2$ for $n>5$ so $0<\dfrac{n}{2^n}<\dfrac{n}{n^2}=\dfrac1n$ for sufficiently large $n$ and by the squeeze theorem from $0,1/n\to0$ it follows $n/2^n\to0$.
Of course, this relies on knowing $1/n\to0$. Do you have that result? If not it might be best to take a more direct epsilon-delta approach.
A: Since $\sum\limits_{n=1}^{\infty}\frac{n}{2^n}$ converges (use for example ratio test), we have $\lim_{n\to\infty}\frac{n}{2^n} = 0$.
A: Here are the steps
$$\lim_{n\to\infty} \cos{\left(\frac{n}{2^{n}}\right)}= \cos{\left( \lim_{n\to\infty} \frac{n}{2^{n}}\right)} $$
$$=\cos\left(\lim_{n\to\infty}\frac{\frac{d}{dn}n}{\frac{d}{dn}2^{n}}\right)=\cos\left(\lim_{n\to\infty}\frac{1}{2^{n}\ln 2}\right) $$
$$=\cos(0)=1$$
A: If $x_n=\frac{n}{2^n}$,
$$\left|\frac{x_{n+1}}{x_n}\right|=\left|\frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}}\right|=\left|\frac{(n+1)2^n}{n 2^{n+1}}\right|=\left|\frac{n+1}{n}\cdot \frac{1}{2}\right|\underset{n\to\infty }{\longrightarrow}  \frac{1}{2}<1,$$
and thus $$\lim_{n\to\infty }\frac{n}{2^n}=0.$$
