MHT CET · Maths · Limits
\(\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}=\)
- A \(0\)
- B \(1\)
- C e
- D \(\frac{1}{\mathrm{e}}\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}=e^{\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}-1\right) \operatorname{cosec} x} \\ & =e^{\lim _{x \rightarrow 0}\left\{\frac{\tan x-\sin x}{1+\sin x} \cdot \frac{1}{\sin x}\right\}} \\ & \left.=e^{\lim _{x \rightarrow 0}\left\{\frac{\frac{1}{\cos x}-1}{1+\sin x}\right.}\right\} \\ & =\mathrm{e}^0=1 \\ & \end{aligned}\)
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