AP EAMCET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}=\)
- A \(\frac{-1}{4}\)
- B \(\frac{1}{2}\)
- C 1
- D 2
Answer & Solution
Correct Answer
(D) 2
Step-by-step Solution
Detailed explanation
\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}…
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