Q86 / 180 · COMEDK 2014 · Mathematics

$\lim _{x \rightarrow 0} \frac{x a^{x}-x}{1-\cos x}$ is equal to

  • A
    $\log a$
  • B
    $\frac{l}{2} \log a$
  • Correct answer
    C
    $2 \log a$
  • D
    $2 \log 2$

Solution

$\lim _{x \rightarrow 0} \frac{x a^{x}-x}{1-\cos x}=\lim _{x \rightarrow 0} \frac{x\left(a^{x}-1\right)}{2 \sin ^{2} \frac{x}{2}}$ $\left[\cos 2 \theta=1-2 \sin ^{2} \theta\right]$ $=\lim _{x \rightarrow 0} \frac{a^{x}-1}{\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}}$ $=\frac{\lim _{x \rightarrow 0} 2\left(\frac{a^{x}-1}{x}\right)}{\left[\lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}}\right]^{2}}=\text{2log a}$

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

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