# Limit $\lim_{n\to\infty}n\frac{\sin\frac{1}{n}-\frac{1}{n}}{1+\frac{1}{n}}$

I need to find a limit of a sequence: $$\lim_{n\to\infty}n\frac{\sin\frac{1}{n}-\frac{1}{n}}{1+\frac{1}{n}}$$

I tried to divide numerator and denominator by n, but it didn't help, as the limit became $\frac{0}{0}$. I tried other things, but always got an indefinite limit. I know that the limit is 0, but I just don't know how to show it. It's probably something really simple, but I'm totally stuck.

• Hint: it doesn't go to 0. – user80944 Dec 2 '13 at 21:37
• Oh, I'm sorry, I put it in wrong! I'll edit it! – Robert Dec 2 '13 at 21:38
• Is this not -1? – Alec Teal Dec 2 '13 at 21:39
• The bottom seems irrelevant, it may still be typed incorrectly. I would use Maclaurin series, though L'Hospital's Rule also works well. – André Nicolas Dec 2 '13 at 21:42
• The whole numerator, sorry if I put it in wrong – Robert Dec 2 '13 at 21:42

Do you know about $\lim\limits_{x\to 0}\frac 1 x \sin (x)$ ?
We have $$\lim_{n\to\infty}n\frac{\sin\frac{1}{n}-\frac{1}{n}}{1+\frac{1}{n}} = \lim_{x\to 0}\frac{\frac{\sin x}{x} - 1}{1+x}$$ Using the fact that $\lim_{x\to 0}\frac{\sin x}{x}=1$, we get the result $$\lim_{x\to 0}\frac{\frac{\sin x}{x} - 1}{1+x}=\frac{1-1}{1}=0$$