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Calculate the following limit:

$$\lim_{x\rightarrow 0} \frac{\sin(\sin(2x))}{x}$$

I'm not allowed to use L'Hôpital's rule. I tried to make the substitutions $2x=y$ and $\sin(2x)=y$, but I couldn't get something to use $\lim_{x\rightarrow 0} \frac{\sin(x)}{x}$ (this is the only one I can use this moment). I tried to use $\sin(2x)=2\sin(x)\cos(x)$ as well, but I didn't get anywhere too. I'm only asking a hint this moment! Thanks!

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    $\begingroup$ Not able or not allowed ? $\endgroup$
    – user65203
    Commented Mar 21, 2020 at 16:05
  • $\begingroup$ allowed. Sorry for my english $\endgroup$
    – Curious
    Commented Mar 21, 2020 at 16:14
  • $\begingroup$ Don't worry, @Dunck. A lot of native English users confuse the two, as well! $\endgroup$
    – amWhy
    Commented Mar 21, 2020 at 16:15
  • $\begingroup$ No problem, but I needed to know. That makes different answers. $\endgroup$
    – user65203
    Commented Mar 21, 2020 at 16:15

1 Answer 1

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We know that $\lim_{x \to 0} \frac{x}{\sin{x}} = 1$, so

$$\lim_{x\rightarrow 0} \frac{\sin(\sin(2x))}{x}= 2\lim_{x\rightarrow 0} \frac{\sin(\sin(2x))}{\sin(2x)}\frac{\sin(2x)}{2x}=2$$

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  • $\begingroup$ Thanks, very helpful! $\endgroup$
    – Curious
    Commented Mar 21, 2020 at 16:09
  • $\begingroup$ Please learn to format trig function: \tan x, \sin x, \cos x, \cot x,\sec x, \csc x` ... $\endgroup$
    – amWhy
    Commented Mar 21, 2020 at 16:12
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    $\begingroup$ @amWhy I just saw your revision and will try to remember, thank you! $\endgroup$
    – Andronicus
    Commented Mar 21, 2020 at 16:12
  • $\begingroup$ No problem, @Andronicus. formatting can sometimes be a steep learning curve! $\endgroup$
    – amWhy
    Commented Mar 21, 2020 at 16:14

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