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

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

Solution

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} \\ & =\lim _{x \rightarrow 0} \frac{2\left(\frac{\sin ^2 x}{x^2}\right) \times(3+\cos x)}{4\left(\frac{\tan 4 x}{4 x}\right)} \\ & =\frac{2(1)^2 \times(3+1)}{4}=2\end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 1)

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