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$\lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2}\left(1-\cos ^2 x\right)}}{x}$ is equal to
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The correct answer is:
$\frac{-1}{\sqrt{2}}$
Let $\begin{aligned} L & =\lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2}\left(1-\cos ^2 x\right)}}{x} \\ & =\lim _{x \rightarrow 0^{-}} \frac{\frac{1}{\sqrt{2}} \sqrt{1-\left(1-\sin ^2 x\right)}}{x} \\ & =\lim _{x \rightarrow 0^{-}} \frac{1}{\sqrt{2}} \frac{|\sin x|}{x}=\frac{1}{\sqrt{2}} \lim _{x \rightarrow 0^{-}} \frac{-\sin x}{x} \\ & =-\frac{1}{\sqrt{2}}\end{aligned}$
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