Q100 / 160 · AP EAMCET 2024 (20 May Shift 1) · Mathematics

$\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{\left(\cos ^{-1} x\right)^2}=$

  • Correct answer
    A
    $-\frac{1}{4}$
  • B
    $\frac{1}{2}$
  • C
    $-\frac{1}{2}$
  • D
    $\frac{1}{4}$

Solution

Let $x=\cos \theta \Rightarrow x \rightarrow 1 \Rightarrow \theta \rightarrow 0$ $\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{\left(\cos ^{-1} x\right)^2}=\lim _{\theta \rightarrow 0} \frac{(\sqrt{\cos \theta}-1)}{\theta^2}$ $=\lim _{\theta \rightarrow 0} \frac{(\cos \theta-1)}{\theta^2(\sqrt{\cos \theta}+1)}=\frac{-2 \sin ^2(\theta / 2)}{\theta^2(2)}=-\frac{1}{4}$

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

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