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$\lim _{x \rightarrow 0} \frac{x a^{x}-x}{1-\cos x}$ is equal to
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Verified Answer
The correct answer is:
$2 \log a$
$\lim _{x \rightarrow 0} \frac{x a^{x}-x}{1-\cos x}=\lim _{x \rightarrow 0} \frac{x\left(a^{x}-1\right)}{2 \sin ^{2} \frac{x}{2}}$
$\left[\cos 2 \theta=1-2 \sin ^{2} \theta\right]$
$=\lim _{x \rightarrow 0} \frac{a^{x}-1}{\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}}$
$=\frac{\lim _{x \rightarrow 0} 2\left(\frac{a^{x}-1}{x}\right)}{\left[\lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}}\right]^{2}}=\text{2log a}$
$\left[\cos 2 \theta=1-2 \sin ^{2} \theta\right]$
$=\lim _{x \rightarrow 0} \frac{a^{x}-1}{\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}}$
$=\frac{\lim _{x \rightarrow 0} 2\left(\frac{a^{x}-1}{x}\right)}{\left[\lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}}\right]^{2}}=\text{2log a}$
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