If \(f(x)\) is the integral of \(\dfrac{1+\cos^{2}x-2\cos x}{x\sin^{2}x\tan x}\), then what is \(\displaystyle\lim_{x\to 0}f'(x)\) equal to?
\(1/4\)
The numerator \(1+\cos^{2}x-2\cos x=(1-\cos x)^{2}\), so \(f'(x)=\dfrac{(1-\cos x)^{2}}{x\sin^{2}x\tan x}\). As \(x\to 0\), using \(1-\cos x\sim \dfrac{x^{2}}{2}\), \(\sin x\sim x\) and \(\tan x\sim x\), the expression behaves like \(\dfrac{x^{4}/4}{x^{4}}=\dfrac{1}{4}\), so the limit is \(1/4\).
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