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$\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{x^{2}}$ is equal to
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Verified Answer
The correct answer is:
$\frac{b^{2}-a^{2}}{2}$
$\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{x^{2}}$
$=\lim _{x \rightarrow 0} \frac{-a \sin a x+b \sin b x}{2 x}$ (using L'Hospital's rule)
$=\lim _{x \rightarrow 0} \frac{-a^{2} \cos a x+b^{2} \sin b x}{2}$
$=\frac{b^{2}-a^{2}}{2}$ (using L'Hospital's rule)
$=\lim _{x \rightarrow 0} \frac{-a \sin a x+b \sin b x}{2 x}$ (using L'Hospital's rule)
$=\lim _{x \rightarrow 0} \frac{-a^{2} \cos a x+b^{2} \sin b x}{2}$
$=\frac{b^{2}-a^{2}}{2}$ (using L'Hospital's rule)
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