Q99 / 160 · AP EAMCET 2024 (18 May Shift 1) · Mathematics

$\lim _{x \rightarrow 0} \frac{1-\cos x \cos 2 x}{\sin ^2 x}=$

  • A
    $\frac{11}{4}$
  • Correct answer
    B
    $\frac{5}{2}$
  • C
    $3$
  • D
    $5$

Solution

Since, $\lim _{x \rightarrow 0} \frac{1-\cos x \cos 2 x}{\sin ^2 x}$ $\begin{aligned} & =\lim _{x \rightarrow 0} \frac{1-\cos x\left(\cos ^2 x-\sin ^2 x\right)}{\sin ^2 x} \\ & =\lim _{x \rightarrow 0}\left(\frac{1-\cos ^3 x}{\sin ^2 x}+\frac{\cos x \sin ^2 x}{\sin ^2 x}\right)\end{aligned}$ $\begin{aligned} & =\lim _{x \rightarrow 0}\left(\frac{1+\cos x+\cos ^2 x}{1+\cos x}+\cos x\right) \\ & =\frac{1+1+1}{1+1}+1=\frac{3}{2}+1=\frac{5}{2}\end{aligned}$

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

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