How to solve this limit?
$$\lim_{x\to 0} \frac{\sin(x\sin(2x))}{x^2}$$
I tried to solve with trigonometric formulas but no result yet. It seems that need to use this $$ \lim_{x\to 0} \frac{\sin(x)}{x} $$
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2Seems like L' Hopital Rule works here. – Mar 16 '21 at 17:09
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In case you are not aware that $\lim_{x\to 0}\frac{\sin x}{x}=1$, then check this. – V.G Mar 16 '21 at 17:11
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$$\frac{\sin(x\sin 2x)}{x^2} = \frac{\sin (x\sin 2x)}{x\sin 2x} \times \frac{\sin 2x}{x} \to 1\times \frac{\sin 2x}{2x} \times 2 \to 1\times 1\times 2 =2 $$
Vishu
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2@Student1058 I think this trick is probably the most common one in evaluating limits involving sines. – Vishu Mar 16 '21 at 17:11
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Note that this depends on the fact that both $x$ and $\sin x$ are $\neq 0$ as $x \to 0$. – Ben Mar 16 '21 at 17:41
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Since $\sin t=t+o(t)$, you have $$ x\sin2x=2x^2+o(x^2) $$ and therefore $$ \sin(x\sin2x)=2x^2+o(x^2) $$ so your limit becomes $$ \lim_{x\to0}\frac{2x^2+o(x^2)}{x^2}=2 $$
egreg
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