<s>[INST] How to tackle $\lim_{x \to 0} \frac{\sin x -\arctan x}{x^2\log(1+x)}$ I am stuck with the following problem that says:

Evaluate : $$\lim_{x \to 0} \frac{\sin x -\arctan x}{x^2\log(1+x)}$$

I tried to use l'Hospital rule to tackle the problem but could not end it.
I will be helpful if someone explain it.Thanks in advance for your time.
 [/INST] \begin{align*}
&\lim_{x\rightarrow 0}\dfrac{\sin x-\tan^{-1}x}{x^{2}\log(1+x)}\\
&=\lim_{x\rightarrow 0}\dfrac{\cos x-1/(1+x^{2})}{2x\log(1+x)+x^{2}/(1+x)}\\
&=\lim_{x\rightarrow 0}\dfrac{(1+x^{2})(1+x)\cos x-(1+x)}{2x(1+x^{2})(1+x)\log(1+x)+x^{2}(1+x^{2})}\\
&=\lim_{x\rightarrow 0}\dfrac{-(1+x^{2})(1+x)\sin x+2x(1+x)\cos x+(1+x^{2})\cos x-1}{2x(1+x^{2})(1+x)(1/(1+x))+(8x^{3}+6x^{2}+4x+2)\log(1+x)+2x+4x^{3}}\\
&=\lim_{x\rightarrow 0}\dfrac{-(x^{3}+x^{2}+x+1)\sin x+(3x^{2}+2x+1)\cos x-1}{6x^{3}+4x+(8x^{3}+6x^{2}+4x+2)\log(1+x)}\\
&=\lim_{x\rightarrow 0}\dfrac{-(3x^{2}+2x+1)\sin x-(x^{3}+x^{2}+x+1)\cos x+(6x+2)\cos x-(3x^{2}+2x+1)\sin x}{18x^{2}+4+(24x^{2}+12x+4)\log(1+x)+(8x^{3}+6x^{2}+4x+2)/(1+x)}\\
&=\dfrac{-1+2}{4+2}\\
&=\dfrac{1}{6}.
\end{align*}
</s>