How do you calculate this limit $$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\sin x)}}{x}?$$ without derivatives please. Thanks.
How do you calculate this limit $\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\sin x)}}{x}$?
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calculus
real-analysis
trigonometry
limits
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23Hint: ${\displaystyle {\sin(\sin(x)) \over x} = {\sin(\sin(x)) \over \sin(x)} {\sin(x) \over x}}$. – 2011-08-04
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3Or, intuitively, since $\lim\limits_{x\to 0}\frac{\sin(x)}{x}=1$, then $\sin(x)\approx x$ when $x\approx 0$, so you expect $\sin(\sin(x))\approx \sin(x)\approx x$ when $x$ is very close to $0$. – 2011-08-04
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0Thanks Zarrax, is just the trick I needed. : D – 2011-08-04
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0@mathsalomon Since a number of nice answers have been given already, please consider accepting one so that the question shows up as answered in the future. – 2011-08-31