$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}=$

$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}=$
  1. $\frac{-1}{4}$
  2. $\frac{1}{2}$
  3. 1
  4. 2

Solution

$ \begin{aligned} & \text { Let } l=\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} \\ & =\lim _{x \rightarrow 0} \frac{2 \sin ^2 x \cdot(3+\cos x)}{x \tan 4 x} \\ & =2 \cdot \lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} \frac{1}{4} \lim _{x \rightarrow 0} \frac{4 x}{\tan 4 x} \cdot \lim _{x \rightarrow 0}(3+\cos x) \end{aligned} $ According to Sandwich theorem, $ \begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \text { and } \lim _{x \rightarrow 0} \frac{\tan x}{x}=1 \\ & =2 \cdot 1 \cdot \frac{1}{4} \cdot 1 \cdot(3+1)=2 \end{aligned} $

Asked in: AP EAMCET 2017 (26 Apr Shift 1)

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