This question already has an answer here:

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, awllower Jun 17 '13 at 11:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


The limit would be $\infty-\infty$ when $x\to0$.

As Amzoti have said in the comment, we can combine the fractions and get:

\begin{align} \lim_{x\to0}\frac{1}{x}-\frac{1}{\sin x}&=\lim_{x\to0}\frac{\sin x-x}{x\sin x},\text{now we can apply L'Hospital's Rule since the limit would be }\frac{0}{0} \\ \\ &=\lim_{x\to0}\frac{\cos x-1}{\sin x+x\cos x}\\ \\ &=\lim_{x\to0}\frac{-\sin x}{\cos x+\cos x-x\sin x}\\ \\ &=\boxed{0} \end{align}

  • $\begingroup$ Use \sin instead of sin. $\endgroup$ – Pedro Tamaroff Jun 17 '13 at 0:23
  • 3
    $\begingroup$ 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 $-x\cos x$. $\endgroup$ – Cameron Buie Jun 17 '13 at 0:23
  • $\begingroup$ I've corrected it, thanks for the tip! Learned something new in $\LaTeX$ $\endgroup$ – 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 $0$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.