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I am curretly working on the following question:

$$\lim\limits_{x \to 0} \frac {x - \tan^{-1}x} {x\sin x}$$

I was thinking of splitting up the limit such as

$$\lim\limits_{x \to 0} \frac{x}{x \sin x} - \lim\limits_{x \to 0} \frac {\tan^{-1}x}{x \sin x}$$

For the LHS, I was thinking of doing L'Hopital's rule and would get

$$\lim\limits_{x \to 0} \frac{1}{x \cos x+\sin x}$$

I feel as if I'm just going in circles and would like guidance towards heading in the right direction.

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  • $\begingroup$ You can split the limit only if the individual limits after splitting exist. Here $\displaystyle\lim_{x\to 0}\dfrac{x}{x\sin x}$ and $\lim\limits_{x \to 0} \dfrac {\tan^{-1}x}{x \sin x}$ do not exit. $\endgroup$ – Yadati Kiran Mar 13 at 6:21
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$$\lim_{x \to 0}\dfrac{x-\arctan x}{x^2}\cdot\lim_{x \to 0}\dfrac x{\sin x}$$

Now apply L’Hôpital for the first limit.

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Use Taylor expansion: $$\lim\limits_{x \to 0} \frac {x - \tan^{-1}x} {x\sin x}=\lim\limits_{x \to 0} \frac {x - \left(x-\frac{x^3}{3}+\frac{x^5}{5}+O(x^7)\right)} {x\sin x}=\\ \lim\limits_{x \to 0} \frac {x\left(\frac{x}{3}-\frac{x^3}{5}+O(x^5)\right)} {\sin x}=0.$$

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