Q88 / 200 · TS EAMCET 2001 · Mathematics

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

  • A
    $\log 10$
  • Correct answer
    B
    $2 \log 10$
  • C
    $3 \log 10$
  • D
    $4 \log 10$

Solution

$\lim _{x \rightarrow 0} \frac{x\left(10^x-1\right)}{1-\cos x}$ 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, $\begin{aligned} & =\lim _{x \rightarrow 0} \frac{x \cdot 10^x(\log 10)^2+10^x \log 10+10^x \log 10}{\cos x} \\ & =\frac{0+\log 10+\log 10}{\cos 0}=2 \log 10\end{aligned}$

Hence, the correct answer is option: 'B'.

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