limx→1 (2−x)Tanπx2

limx→1 (2−x)tan⁡πx2 1∞ form =elimx→1 tan⁡πx2[2−x−1]=elimx→1 tan⁡πx2[1−x]=elimx→1 1−xcot⁡πx2=elimx→1-1-cosec2⁡πx2π2 L hospital rule=e−1−12π=e2π

limx→1 (2−x)Tanπx2

limx→1 (2−x)tan⁡πx2 1∞ form =elimx→1 tan⁡πx2[2−x−1]=elimx→1 tan⁡πx2[1−x]=elimx→1 1−xcot⁡πx2=elimx→1-1-cosec2⁡πx2π2 L hospital rule=e−1−12π=e2π

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\lim_{x \to 1} {(2 - x)}^{T a n \frac{π x}{2}}$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$e^{\frac{1}{π}}$

$e^{\frac{2}{π}}$

$- e^{\frac{2}{π}}$

$e$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} lim_{x \to 1} (2 - x)^{\frac{\tan π x}{2}} (1^{\infty} form) \\ = e^{lim_{x \to 1} \tan \frac{π x}{2} [2 - x - 1]} \end{array}$

$\begin{array}{l} = e^{lim_{x \to 1} \tan \frac{π x}{2} [1 - x]} \\ = e^{lim_{x \to 1} \frac{1 - x}{\cot \frac{π x}{2}}} \\ = e^{\lim_{x \to 1} \frac{- 1}{- c {osec}^{2} (\frac{π x}{2}) \frac{π}{2}}} L h o s p i t a l r u l e \\ = e^{\frac{- 1}{- 1} \frac{2}{π}} \\ = e^{\frac{2}{π}} \end{array}$