# How do I solve this without using L'Hôpital's?

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

• You could use the taylor series of $sin(x)$ , $cos(3x)$ and $e^x$. – Peter Apr 18 '16 at 22:28
• @Peter Inserting Taylor series and comparing leading terms is exactly the same as using L'Hopital. – 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 $\lim_{x\to0}(\sin x-x)$?
• Beat me to it! $+1$ – Cameron Williams Apr 18 '16 at 22:45