Question:
\(\lim\limits_{x\to0}\left(\tan\left(\frac{\pi}{4}+x\right)\right)^{\frac1x}\) is equal to :
\(\lim\limits_{x\to0}\left(\tan\left(\frac{\pi}{4}+x\right)\right)^{\frac1x}\) is equal to :
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It is a \(1^\infty\) form: the limit is \(e^{\lim (\tan(\pi/4+x)-1)/x}\).
Updated On: Nov 5, 2026
- \(2\)
- \(1\)
- \(e\)
- \(e^2\)
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The Correct Option is D
Solution and Explanation
The limit is of the form \(1^\infty\), so \(L=e^{\ell}\) with \(\ell=\lim_{x\to0}\frac{\tan\left(\frac\pi4+x\right)-1}{x}\).
\(\tan\left(\frac\pi4+x\right)=\frac{1+\tan x}{1-\tan x}\), hence \(\tan\left(\frac\pi4+x\right)-1=\frac{2\tan x}{1-\tan x}\).
\(\ell=\lim_{x\to0}\frac{2\tan x}{x(1-\tan x)} = 2\cdot1\cdot\frac{1}{1-0}=2\).
So \(L=e^2\).
Answer: (4) \(e^2\)
\(\tan\left(\frac\pi4+x\right)=\frac{1+\tan x}{1-\tan x}\), hence \(\tan\left(\frac\pi4+x\right)-1=\frac{2\tan x}{1-\tan x}\).
\(\ell=\lim_{x\to0}\frac{2\tan x}{x(1-\tan x)} = 2\cdot1\cdot\frac{1}{1-0}=2\).
So \(L=e^2\).
Answer: (4) \(e^2\)
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