MHT CET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}\) has the value
- A 2
- B \(\frac{1}{2}\)
- C 4
- D 3
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} \\ & =\lim _{x \rightarrow 0} \frac{2\left(\frac{\sin ^2 x}{x^2}\right) \times(3+\cos x)}{4\left(\frac{\tan 4 x}{4 x}\right)} \\ & =\frac{2(1)^2 \times(3+1)}{4}=2\end{aligned}\)
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