# Evaluate limit $\lim_{x \rightarrow 0}\left (\frac 1x- \frac 1{\sin x} \right )$ [duplicate]

Can someone provide me with some hint how to evaluate this limit? $$\lim_{x \rightarrow 0}\left (\frac 1x- \frac 1{\sin x} \right )$$ I tried l'hopital's rule but it didn't work.

## marked as duplicate by Martin Sleziak, Lord_Farin, Stefan Hansen, Julian Kuelshammer, awllowerJun 17 '13 at 11:45

The limit would be $\infty-\infty$ when $x\to0$.
• Use \sin instead of sin. – Pedro Tamaroff Jun 17 '13 at 0:23
• If you put $\sin x$ instead of $sin x$, you'll get the more "operator-looking" $\sin x$. Also, you should have $-x\sin x$ in your last denominator, rather than $-x\cos x$. – Cameron Buie Jun 17 '13 at 0:23
• I've corrected it, thanks for the tip! Learned something new in $\LaTeX$ – user67258 Jun 17 '13 at 0:25
Rewrite the difference as (after Maclaurin series expansion and some algebra) $$\frac{\sin x-x}{x \sin x}=\frac{x+O(x^3)-x}{x(x+O(x^3))} =\frac{O(x)}{1+O(x^2)}$$ which tends to $0$.