Q102 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics

$\cos \left[\lim _{x \rightarrow \infty} \frac{2 \pi|x|+\pi x}{|x|-3 x}+\lim _{x \rightarrow 0} \frac{\cos \left(\frac{\pi}{2} \cos ^2 x\right)}{x^2}\right]$

  • A
    1
  • Correct answer
    B
    -1
  • C
    0
  • D
    $\frac{1}{\sqrt{2}}$

Solution

$\begin{aligned} & \text { } \cos \left[\lim _{x \rightarrow \infty} \frac{2 \pi|x|+\pi x}{|x|-3 x}+\lim _{x \rightarrow 0} \frac{\cos \left(\frac{\pi}{2} \cos ^2 x\right)}{x^2}\right] \\ & =\cos \left[\lim _{x \rightarrow \infty} \frac{2 \pi x+\pi x}{x-3 x}+\lim _{x \rightarrow 0} \frac{\cos \left(\frac{\pi}{2}\left(1-\sin ^2 x\right)\right)}{x^2}\right] \\ & =\cos \left[\lim _{x \rightarrow \infty} \frac{3 \pi x}{-2 x}+\lim _{x \rightarrow 0}\left(\frac{\sin \left(\frac{\pi}{2} \sin ^2 x\right)}{\left(\frac{\pi}{2} \sin ^2 x\right)} \times \frac{\frac{\pi}{2} \sin ^2 x}{x^2}\right)\right] \\ & =\cos \left[-\frac{3 \pi}{2}+\frac{\pi}{2}\right]=\cos (-\pi)=-1 .\end{aligned}$

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

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