lim _{x rightarrow 0} left( frac{x cdot 10^x | vedclass

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(B) Given limit: $L = \lim _{x \rightarrow 0} \frac{x(10^x - 1)}{1 - \cos x}$.Since this is a $\frac{0}{0}$ form,we apply $L$-Hospital's rule:$L = \li

Vedclass · Accepted Answer

$2 \log 10$

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(B) Given limit: $L = \lim _{x \rightarrow 0} \frac{x(10^x - 1)}{1 - \cos x}$.Since this is a $\frac{0}{0}$ form,we apply $L$-Hospital's rule:$L = \lim _{x \rightarrow 0} \frac{x \cdot 10^x \ln 10 + (10^x - 1)}{\sin x}$.Again,this is a $\frac{0}{0}$ form,so we apply $L$-Hospital's rule again:$L = \lim _{x \rightarrow 0} \frac{x \cdot 10^x (\ln 10)^2 + 10^x \ln 10 + 10^x \ln 10}{\cos x}$.Substituting $x = 0$:$L = \frac{0 \cdot 10^0 (\ln 10)^2 + 10^0 \ln 10 + 10^0 \ln 10}{\cos 0} = \frac{0 + \ln 10 + \ln 10}{1} = 2 \ln 10$.Since $\log 10$ usually denotes $\log_{10} 10 = 1$ or $\ln 10$ depending on context,and given the options,the result is $2 \log 10$.

$\lim _{x \rightarrow 0} \left( \frac{x \cdot 10^x - x}{1 - \cos x} \right)$ is equal to

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