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$\lim _{x \rightarrow 0} \frac{x^4+x^3+x^2}{\sin ^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) \cdot \tan ^{-1} x}=$
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Let $L=\lim _{x \rightarrow 0} \frac{x^4+x^3+x^2}{\sin ^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) \tan ^{-1} x}$
$\begin{aligned} & \quad=\lim _{x \rightarrow 0} \frac{x^4+x^3+x^2}{\left(\tan ^{-1} x\right)^2}=\lim _{x \rightarrow 0} \frac{4 x^3+3 x^2+2 x}{2 \tan ^{-1} x \cdot \frac{1}{1+x^2}} \\ & =\lim _{x \rightarrow 0} \frac{\left(4 x^3+3 x^2+2 x\right)\left(1+x^2\right)}{2 \tan ^{-1} x} \\ & =\lim _{x \rightarrow 0} \frac{\left(12 x^2+6 x+2\right)\left(1+x^2\right)+\left(4 x^3+3 x^2+2 x\right)(2 x)}{\left(\frac{2}{1+x^2}\right)} \\ & =\frac{2}{2}=1\end{aligned}$
$\begin{aligned} & \quad=\lim _{x \rightarrow 0} \frac{x^4+x^3+x^2}{\left(\tan ^{-1} x\right)^2}=\lim _{x \rightarrow 0} \frac{4 x^3+3 x^2+2 x}{2 \tan ^{-1} x \cdot \frac{1}{1+x^2}} \\ & =\lim _{x \rightarrow 0} \frac{\left(4 x^3+3 x^2+2 x\right)\left(1+x^2\right)}{2 \tan ^{-1} x} \\ & =\lim _{x \rightarrow 0} \frac{\left(12 x^2+6 x+2\right)\left(1+x^2\right)+\left(4 x^3+3 x^2+2 x\right)(2 x)}{\left(\frac{2}{1+x^2}\right)} \\ & =\frac{2}{2}=1\end{aligned}$
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