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!
2026-04-04 14:47:20.1775314040
How do I solve this without using L'Hôpital's?
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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 …
Hints.
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).
What is $\lim_{x\to0}(\sin x-x)$?