Task: $$\lim _{x\to 0}\left(\frac{\sin\left(2x\right)-2\sin\left(x\right)}{x\cdot \:arctan^2x}\right)$$

Hello! It is necessary to solve the next limit. I try several methods. For example, if you use the equivalent of small, it turns out: $$\sin\left(2x\right)\rightarrow 2x$$ $$2\sin\left(x\right)\rightarrow 2x$$ $$x\cdot arc\tan^2\left(x\right)\rightarrow x\cdot x^2$$ Limit obtained indefinite again...(

I have tried to simplify the numerator (only in the numerator as the denominator have no idea what to do): $$\lim _{x\to 0}\left(\frac{2\sin x\cdot \cos x-2\sin x}{x\cdot arc\tan^2x}\right)=\lim _{x\to 0}\left(\frac{\sin x\left(\cos x-2\right)}{x\cdot arc\tan^2x}\right)=?$$

  • $\begingroup$ what is $arctg^2x$? $\endgroup$ – Varun Iyer Nov 15 '15 at 14:12
  • $\begingroup$ @VarunIyer I think it's another notation for $\arctan^2x$. $\endgroup$ – Workaholic Nov 15 '15 at 14:13
  • $\begingroup$ $\arctan ^2\left(x\right)$ $\endgroup$ – do.this Nov 15 '15 at 14:14

Use equivalents:

Numerator: $\;\sin 2x-2\sin x=2\sin x(\cos x-1)$, and $\sin x\sim_0 x$, $\;\cos x-1\sim_0 -\dfrac{x^2}2$.

Denominator: $\;\arctan x\sim_0 x$.

Thus: $$\frac{\sin 2x-2\sin x}{x\arctan^2 x}\sim_0\frac{x(-x^2)}{x\cdot x^2}=-1.$$

  • $\begingroup$ So the answer must be $-1$ $\endgroup$ – do.this Nov 15 '15 at 14:28
  • $\begingroup$ I beg you pardon? Did I miss something? $\endgroup$ – Bernard Nov 15 '15 at 14:29
  • $\begingroup$ Well, you got $+\infty $. And it is necessary to limit equal $-1$. $\endgroup$ – do.this Nov 15 '15 at 14:31
  • $\begingroup$ Oops! I took a wrong version of the numerator. I'll correct that at once. Thanks for pointing it! $\endgroup$ – Bernard Nov 15 '15 at 14:35
  • $\begingroup$ ooooh, thanks!!! $\endgroup$ – do.this Nov 15 '15 at 14:42


$$\dfrac{\sin2x-2\sin x}{x\arctan^2x}=-2\cdot\dfrac{\sin x}x\cdot\dfrac{1-\cos x}{x^2}\cdot\left(\dfrac x{\arctan x}\right)^2$$

and $$\dfrac{1-\cos x}{x^2}=\dfrac1{(1+\cos x)}\cdot\left(\dfrac{\sin x}x\right)^2$$

  • $\begingroup$ And all this can now till the end simplify? $\endgroup$ – do.this Nov 15 '15 at 14:27
  • $\begingroup$ $\displaystyle\lim_{x\to 0}\dfrac{1-\cos x}{x^2}=\frac12$ is a basic limit. $\endgroup$ – Bernard Nov 15 '15 at 14:29
  • $\begingroup$ @do.this, You are aware of proofwiki.org/wiki/Limit_of_Sine_of_X_over_X? $\endgroup$ – lab bhattacharjee Nov 15 '15 at 14:31
  • $\begingroup$ @do.this, You miscalculated $$2\sin x\cos x-2\sin x=2\sin x(\cos x-1)$$ $\endgroup$ – lab bhattacharjee Nov 15 '15 at 14:34
  • $\begingroup$ Had in mind, can you paint now all this together with a limit? $\endgroup$ – do.this Nov 15 '15 at 14:36

So you have the following limit:

$$\lim _{x\to 0}\left(\frac{\sin\left(2x\right)-2\sin\left(x\right)}{x\cdot \:\arctan^2x}\right)$$

I would suggest using taylor series:

$$\lim _{x\to 0}\left(\frac{\sin\left(2x\right)-2\sin\left(x\right)}{x\cdot \:\arctan^2x}\right) = \frac{\left(2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \ldots\right) - 2\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\right)}{x\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\right)^2} = \frac{-x^3 + \ldots}{x^3 + \ldots}$$

Note that we only care about the lowest power of $x$ on the numerator and the denominator. In this case it is $x^3$. Thus, if we multiply the fraction by $\frac{1}{x^3}$, we end up with:

$$\frac{-x^3 + \ldots}{x^3 + \ldots} = \frac{-1 + 0 + \ldots}{1 + 0 + \ldots} = -1$$

Therefore our answer is $-1$. Comment if you have questions.


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