The value of limx→1 tan⁡πx4tan⁡πx2 is - Infinity Learn by Sri Chaitanya

A
e^-2
B
e^-1
C
e
D
1

Solution:

$lim_{x \to 0} {(\tan \frac{πx}{4})}^{\tan \frac{πx}{2}}$

$= e^{lim_{x \to 1} (\tan \frac{πx}{4} - 1) \tan \frac{πx}{2}} = e^{lim_{x \to 1} (\frac{\sin \frac{πx}{4} - \cos \frac{πx}{4}}{\cos \frac{πx}{4}}) \frac{2 \sin \frac{πx}{4} \cos \frac{πx}{4}}{\cos \frac{πx}{2}}}$

$= e^{\frac{lim_{x \to 1} 2 \sin^{2} \frac{πx}{4} - 2 \sin \frac{πx}{4} \cos \frac{πx}{4}}{\cos \frac{πx}{2}}} = e^{lim_{x \to 1} (\frac{1 - \cos \frac{πx}{2} - \sin \frac{πx}{2}}{\cos \frac{πx}{2}})}$

$= e^{lim_{x \to 1} \frac{\sin \frac{πx}{2} \times \frac{π}{2} - \cos \frac{πx}{2} \times \frac{π}{2}}{- \sin \frac{πx}{2} \times \frac{π}{2}}} = e^{- 1}$

The value of $lim_{x \to 1} {(\tan \frac{πx}{4})}^{\tan \frac{πx}{2}}$ is

Solution:

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