Finding the limit $\lim\limits_{n\to\infty}n\left[\log_2(n+1)-2\log_2 n-\log_2(\sin \frac{1}{n})\right]$ How could i find following limit.
$\lim\limits_{n\to\infty}n\left[\log_2(n+1)-2\log_2 n-\log_2(\sin \frac{1}{n})\right]$.
I tried $\lim\limits_{n\to\infty}n \log_2\frac{(n+1)}{n^2\sin \frac{1}{n}}$ but i stuck.
 A: You can rewrite this as
$${1\over\ln2}\left(\lim_{n\to\infty}n\ln\left(n+1\over n\right)+\lim_{n\to\infty}n\ln\left(1/n\over\sin(1/n) \right) \right)$$
As others have noted, the first limit is $1$, because $\left(1+{1\over n}\right)^n\to e$ as $n\to\infty$.  The second limit, however, has to be done a little bit carefully.  It's not enough to note that $(1/n)/\sin(1/n)\to1$ in order to conclude that $\ln1=0$ implies the limit is $0$; that same logic would imply the first limit is $0$ as well.
Letting $x=1/n$, the second limit can be rewritten as
$$\lim_{x\to0}{\ln x-\ln\sin x\over x}$$
L'Hopital says this should equal
$$\lim_{x\to0}\left({1\over x}-{\cos x\over\sin x}\right)=\lim_{x\to0}{\sin x-x\cos x\over x\sin x}$$
and L'Hopital again says this should equal
$$\lim_{x\to0}{x\sin x\over\sin x+x\cos x}=\lim_{x\to0}{\sin x\over{\sin x\over x}+\cos x}={0\over1+1}=0$$
Putting everything together, the desired limit is 
$${1\over\ln2}(1+0)={1\over\ln2}$$
A: $\lim\limits_{n\to\infty}n\left[\log_2(n+1)-2\log_2 n-\log_2(\sin \frac{1}{n})\right]$.
$\lim\limits_{n\to\infty} \log_2(\frac{(n+1)}{n^2\sin \frac{1}{n}})^n$
$\lim\limits_{n\to\infty} \log_2(\frac{(1+1/n)}{(\frac{\sin \frac{1}{n}}{1/n})})^n$
when n goes to infinity $1/n$ goes to zero
then $(\frac{\sin \frac{1}{n}}{1/n})$ goes to $1$
$\lim\limits_{n\to\infty} \log_2(\frac{(1+0)}{1})^n$
$\lim\limits_{n\to\infty} \log_21^n$
when $n=1$, $1^n=1$
the answer is $0$
A: So like Barry said we have:
$${1\over\log 2}\left(\lim_{n\to\infty}n\log\left(n+1\over n\right)-\lim_{n\to\infty}n\log\left(\sin(1/n) \over 1/n \right) \right)$$
An alternative way of showing the second limit tends to $0$:
We will use the fact that $x - \dfrac {x^3} {3!}\le \sin x \le x$ for $x \ge 0$.
It follows that:
$$n \log \left(1 - \frac 1 {3! n^2} \right)\le n \log \left( \frac {\sin (1/n)} {1/n}\right) \le 0$$
If we can show the LHS tends to $0$ we have the desired result by squeeze/sandwhich theorem.
Well $$\lim_{n \to \infty} n \log \left(1 - \frac 1 {3! n^2} \right) = \lim_{n \to \infty}\frac 1 {3!n}\left[3!n^2 \log \left( 1 - \ \frac 1 {3! n^2} \right)\right] = -\frac 1 {3!} \lim_{n \to \infty} \frac 1 n=0$$
A: We will use $\lim\limits_{x\to0}\color{#C00000}{\frac{\log(1+x)}x}=\color{#C00000}{1}$ and $\lim\limits_{x\to0}\color{#00A000}{\frac{\sin(x)-x}{x^2}}=\color{#00A000}{0}$.
$$
\begin{align}
&\lim_{n\to\infty}n\left[\log_2(n+1)-2\log_2(n)-\log\left(\sin\left(\tfrac1n\right)\right)\right]\\
&=\frac1{\log(2)}\lim_{n\to\infty}\color{#C00000}{\frac{\log\left(1+\frac1n\right)}{\frac1n}}
-\log_2\left[\lim_{n\to\infty}\left(1+\color{#00A000}{\frac{\sin\left(\frac1n\right)-\frac1n}{\frac1{n^2}}}\frac1n\right)^n\,\right]\\
&=\frac1{\log(2)}\cdot\color{#C00000}{1}-\log_2\left(e^{\color{#00A000}{0}}\right)\\[3pt]
&=\frac1{\log(2)}
\end{align}
$$
A: One has $$n\log_2 \frac{n+1}{n^2sin(\frac 1n)}$$ Make $n=\frac1x$ so $x\to 0$ this becomes by continuity of $\log_2$ and the well known limit $\frac{x}{sin x}$
$$\frac 1x \log_2\frac{(1+x)x}{sin x}\to\frac1x\log_2(1+x)\to\log_2(1+x)^{\frac1x}\to\log_2 e$$
The answer is $\log_2 e$ where $e$ is the base of natural logaritms
