Without L'Hopital,

$$\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$$

This is

$$\frac{\sin x -\frac{\sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\frac{\sin x \cdot \cos x - \sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\sin x\cdot \cos x - \sin x}{x^2\cdot\sin 2x\cdot \cos x}$$

Split that:

$$\frac{\sin x\cdot \cos x}{x^2\cdot\sin 2x\cdot \cos x} - \frac{\sin x}{x^2\cdot\sin 2x\cdot \cos x}$$

In the left side, we can cancel the $\cos x$ and also apply $\frac{\sin x}{x} = 1$ once:

$$\frac{1}{x\cdot\sin 2x} - \frac{\sin x}{x^2\cdot\sin 2x\cdot \cos x}$$

That was probably a bad idea, since $x \cdot \sin2x$ will definitely be $0$... But anyway, let's keep going with the right side. There, we can apply the identity $\frac{\sin x}{x} = 1$ again:

$$\frac{1}{x\cdot\sin 2x} - \frac{1}{x\cdot\sin 2x\cdot \cos x}$$

Hey, I could get rid of the $\sin 2x$ on the left side if I multiply and divide by $2x$... the same on the right side:

$$\frac{1}{2x^2} - \frac{1}{2x^2\cdot \cos x}$$

Looking pretty, but sadly that's not going anywhere. What can I do?


$$\frac{\sin x-\tan x}{x^{2}\sin 2x}=\frac{\sin x}{\sin 2x}\frac{\cos x -1}{x^{2} \cos x}=\frac{\sin x}{\sin 2x}\frac{-2\sin^{2}\frac{x}{2}}{x^{2}\cos x}$$ Now, $$\frac{\sin x}{\sin 2x}\to \frac{1}{2}$$ $$\frac{\sin^{2}\frac{x}{2}}{x^{2}}=1/4\frac{\sin^{2}\frac{x}{2}}{(x/2)^{2}}\to 1/4$$ And $\cos x \to 1$ So the limit is $-1/4$.

  • $\begingroup$ How come $\frac{\sin x}{\sin 2x}\to \frac{1}{2}$? And also, I'm not sure I quite grasped the last line. I know that $\cos x \to 1$, but what does that do anyway? $\endgroup$ – Zol Tun Kul Oct 1 '15 at 8:34
  • 1
    $\begingroup$ You can look at it as $\frac{1}{2} \frac{\sin x}{x} \frac{2x}{\sin 2x}$ $\endgroup$ – preferred_anon Oct 1 '15 at 8:38
  • $\begingroup$ You need to know $\cos(x) \to 1$ because it appears in the fraction at the end of the first line. $\endgroup$ – preferred_anon Oct 1 '15 at 8:39

$$\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x} =-\lim_{x\to0}\frac{\sin x(1-\cos x)}{2x^2\sin x\cos^2x} =-\lim_{x\to0}\frac{(1-\cos x)}{2x^2}\cdot\frac1{\lim_{x\to0}\cos^2x}$$

Now $$\lim_{x\to0}\frac{(1-\cos x)}{2x^2}=\lim_{x\to0}\frac{(1-\cos x)(1+\cos x)}{2x^2}\cdot\dfrac1{\lim_{x\to0}(1+\cos x)}=?$$

  • $\begingroup$ It appears I am lacking on trigonometric properties. How did you go from $(\sin x - \tan x)$ to $\sin x(1 - \cos x)$? $\endgroup$ – Zol Tun Kul Oct 1 '15 at 8:10
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    $\begingroup$ Did you just write $1/0$, disguised as $1/( \lim_{x \to 0} \sin x)$? $\endgroup$ – Najib Idrissi Oct 1 '15 at 8:10
  • $\begingroup$ @ZolTunKul, $$\sin x-\tan x=\dfrac{\sin x(\cos x-1)}{\cos x}$$ $\endgroup$ – lab bhattacharjee Oct 1 '15 at 8:11
  • $\begingroup$ @ZolTunKul, Please find updated answer $\endgroup$ – lab bhattacharjee Oct 1 '15 at 8:12

Notice, $$\lim_{x\to 0}\frac{\sin x-\tan x}{x^2\sin 2x}$$ $$=\lim_{x\to 0}\frac{\sin x\frac{(\cos x-1)}{\cos x}}{2x^2\sin x\cos x}$$ $$=\frac{1}{2}\lim_{x\to 0}\frac{\cos x-1}{x^2\cos^2 x}$$ $$=\frac{1}{2}\lim_{x\to 0}\frac{\cos x-1}{x^2}\cdot \lim_{x\to 0}\frac{1}{\cos^2x}$$ $$=\frac{1}{2}\lim_{x\to 0}\frac{\left(1-\frac{x^2}{2!}+O(x^2)\right)-1}{x^2}\cdot 1$$ $$=\frac{1}{2}\lim_{x\to 0}\frac{\left(-\frac{x^2}{2!}+O(x^2)\right)}{x^2}$$ $$=\frac{1}{2}\lim_{x\to 0}\left(-\frac{1}{2!}+O(1)\right)$$ $$=\frac{1}{2}\left(-\frac{1}{2}+0\right)=\color{red}{-\frac{1}{4}}$$


first :

$$\lim_{x\to 0}\frac{\sin x-\tan x}{x^3}=\frac{-1}{2}$$

proof :$\lim_{x\to 0} \frac{\tan ^nx - \sin ^m x}{x^r}=?$without l'Hôpital's rule.

now :

$$\lim_{x\to 0}\frac{\sin x-\tan x}{x^2\sin 2x}=\lim_{x\to 0}\frac{\sin x-\tan x}{x^3}.\frac{2x}{2\sin 2x}=?$$

since :

$$\lim_{x\to 0}\frac{2x}{\sin 2x}=1$$

so :

$$\lim_{x\to 0}\frac{\sin x-\tan x}{x^2\sin 2x}=\frac{-1}{2}.\frac{1}{2}=\frac{-1}{4}$$


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