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