Ltx→12 sin πxcos πx2e2x-2e=

Ltx→12 sin πxcos πx2e2x-2e =Ltx→12 2sin πxcos πx4ee2x-1-1 =Ltx→12 sin 2πx4ee2x-1-1 =Ltx→12 sin (π-2πx)4ee2x-1-1 =Ltx→12 sin π(2x-1)2x-14ee2x-1-12x-1 =-π4e

Ltx→12 sin πxcos πx2e2x-2e=

Ltx→12 sin πxcos πx2e2x-2e =Ltx→12 2sin πxcos πx4ee2x-1-1 =Ltx→12 sin 2πx4ee2x-1-1 =Ltx→12 sin (π-2πx)4ee2x-1-1 =Ltx→12 sin π(2x-1)2x-14ee2x-1-12x-1 =-π4e

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

$\underset{x \to \frac{1}{2}}{Lt} \frac{\sin π x \cos π x}{2 e^{2 x} - 2 e} =$

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

$- \frac{π}{4 e}$

$\frac{π}{4 e}$

$- \frac{1}{4 π e}$

$\frac{1}{4 π e}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\underset{x \to \frac{1}{2}}{Lt} \frac{\sin π x \cos π x}{2 e^{2 x} - 2 e} = \underset{x \to \frac{1}{2}}{Lt} \frac{2 \sin π x \cos π x}{4 e (e^{2 x - 1} - 1)} = \underset{x \to \frac{1}{2}}{Lt} \frac{\sin 2 π x}{4 e (e^{2 x - 1} - 1)} = \underset{x \to \frac{1}{2}}{Lt} \frac{\sin (π - 2 π x)}{4 e (e^{2 x - 1} - 1)} = \underset{x \to \frac{1}{2}}{Lt} \frac{\frac{\sin π (2 x - 1)}{2 x - 1}}{4 e (\frac{e^{2 x - 1} - 1}{2 x - 1})} = - \frac{π}{4 e}$