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$\lim _{x \rightarrow 0}\left(\frac{x \cdot 10^x-x}{1-\cos x}\right)$ is equal to
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Verified Answer
The correct answer is:
$2 \log 10$
$\lim _{x \rightarrow 0} \frac{x\left(10^x-1\right)}{1-\cos x}$
Using L-Hospital's rule,
$=\lim _{x \rightarrow 0} \frac{x 10^x \log 10+10^x-1}{\sin x}$
Again using L-Hospital's rule,
$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{x \cdot 10^x(\log 10)^2+10^x \log 10+10^x \log 10}{\cos x} \\ & =\frac{0+\log 10+\log 10}{\cos 0}=2 \log 10\end{aligned}$
Using L-Hospital's rule,
$=\lim _{x \rightarrow 0} \frac{x 10^x \log 10+10^x-1}{\sin x}$
Again using L-Hospital's rule,
$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{x \cdot 10^x(\log 10)^2+10^x \log 10+10^x \log 10}{\cos x} \\ & =\frac{0+\log 10+\log 10}{\cos 0}=2 \log 10\end{aligned}$
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