How can I solve this problem without using L'Hôpital's rule?$$\lim_{x→0}\frac{(\sin(x)-x)(\cos(3x)-1)}{x(e^x -1)}$$

Thanks in advance!

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    $\begingroup$ You could use the taylor series of $sin(x)$ , $cos(3x)$ and $e^x$. $\endgroup$ – Peter Apr 18 '16 at 22:28
  • $\begingroup$ @Peter Inserting Taylor series and comparing leading terms is exactly the same as using L'Hopital. $\endgroup$ – Arthur Apr 18 '16 at 22:46

You can use the two basic limits $$ \lim_{t\to0}\frac{1-\cos t}{t^2}=\frac{1}{2}, \qquad \lim_{x\to0}\frac{e^x-1}{x}=1 $$ so you can rewrite your limit as $$ \lim_{x\to0}-9(\sin x-x)\frac{1-\cos(3x)}{(3x)^2}\frac{x}{e^x-1} $$ and conclude the limit is …


  1. Use the fact that $$ \lim_{x\to x_0}f(x)g(x)= \lim_{x\to x_0}f(x)\cdot\lim_{x\to x_0}g(x) $$ provided both limits on the right hand side exist (which generalizes to a product of three or more factors).

  2. What is $\lim_{x\to0}(\sin x-x)$?

  • $\begingroup$ Beat me to it! $+1$ $\endgroup$ – Cameron Williams Apr 18 '16 at 22:45
  • $\begingroup$ @KonKan “Not helpful enough→no vote”; “fails to communicate information→downvote”. $\endgroup$ – egreg Apr 18 '16 at 23:38
  • $\begingroup$ @egreg: thank you for the clarification. Now, I'm sure we are both busy enough to go on with this! I'm happy you got my point ;) $\endgroup$ – KonKan Apr 18 '16 at 23:42

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