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$\lim _{x \rightarrow 0} \frac{x^2 \log (\cos x)}{\log \left(1+x^2\right)}=$
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$0$
We are given that $\lim _{x \rightarrow 0} \frac{x 2 \log (\cos x)}{\log \left(1+x^2\right)}\left(\frac{0}{0}\right.$ form $)$
$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{x^2\left(\frac{1}{\cos x}\right) \cdot(-\sin x)+\log (\cos x) \cdot 2 x}{\left(\frac{1}{1+x^2}\right) \cdot 2 x} \\ & =\lim _{x \rightarrow 0} \frac{\left[=x^2 \sin x+2 x \cos x \cdot \log (\cos x)\right]\left(1+x^2\right)}{2 x \cos x} \\ & =\lim _{x \rightarrow 0} \frac{x\left(1+x^2\right)[-x \sin x+2 \cos x \log (\cos x)]}{2 x \cos x} \\ & =\lim _{x \rightarrow 0} \frac{\left(1+x^2\right)[-x \sin x+2 \cos x \cdot \log (\cos x)]}{2 \cos x} \\ & =\frac{1 \cdot[0+2 \cdot \log (1)]}{2 \cos \theta}=\frac{0}{2}=0\end{aligned}$
$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{x^2\left(\frac{1}{\cos x}\right) \cdot(-\sin x)+\log (\cos x) \cdot 2 x}{\left(\frac{1}{1+x^2}\right) \cdot 2 x} \\ & =\lim _{x \rightarrow 0} \frac{\left[=x^2 \sin x+2 x \cos x \cdot \log (\cos x)\right]\left(1+x^2\right)}{2 x \cos x} \\ & =\lim _{x \rightarrow 0} \frac{x\left(1+x^2\right)[-x \sin x+2 \cos x \log (\cos x)]}{2 x \cos x} \\ & =\lim _{x \rightarrow 0} \frac{\left(1+x^2\right)[-x \sin x+2 \cos x \cdot \log (\cos x)]}{2 \cos x} \\ & =\frac{1 \cdot[0+2 \cdot \log (1)]}{2 \cos \theta}=\frac{0}{2}=0\end{aligned}$
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