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Question

Mathematics Question on Limits

lim⁡x→12sin⁡(cos⁡−1(x))−x1−tan⁡(cos⁡−1(x))\lim_{{x \to \frac{1}{\sqrt{2}}}} \frac{\sin(\cos^{-1}(x)) - x}{1 - \tan(\cos^{-1}(x))}
is equal to :

A

2\sqrt2

B

−2-\sqrt2

C

12\frac{1}{\sqrt2}

D

−12-\frac{1}{\sqrt2}

Answer

−12-\frac{1}{\sqrt2}

Explanation

Solution

The correct answer is (D) : −12-\frac{1}{\sqrt2}
lim⁡x→12sin⁡(cos⁡−1(x))−x1−tan⁡(cos⁡−1(x))\lim_{{x \to \frac{1}{\sqrt{2}}}} \frac{\sin(\cos^{-1}(x)) - x}{1 - \tan(\cos^{-1}(x))}
let cos−1⁡x=π4+θcos^{−1}⁡x=\frac{π}{4}+θ
lim⁡θ→0sin⁡(π4+θ)−cos⁡(π4+θ)1−tan⁡(π4+θ)\lim_{{\theta \to 0}} \frac{{\sin\left(\frac{\pi}{4} + \theta\right) - \cos\left(\frac{\pi}{4} + \theta\right)}}{{1 - \tan\left(\frac{\pi}{4} + \theta\right)}}
lim⁡θ→02sin⁡(π4+θ−π4)1−1+tan⁡θ1−tan⁡θ\lim_{{\theta \to 0}} \frac{{\sqrt{2}\sin\left(\frac{\pi}{4} + \theta - \frac{\pi}{4}\right)}}{{1 - \frac{1 + \tan\theta}{1 - \tan\theta}}}
lim⁡θ→02sin⁡(θ)−2tan⁡(θ)(1−tan⁡(θ)=−12\lim_{{\theta \to 0}} \frac{{\sqrt{2}\sin(\theta)}}{{-2\tan(\theta)}}(1 - \tan(\theta) = -\frac{1}{\sqrt{2}}