limx→0sin2π2−3xsec2π2−5x=

limx→0sin2π2−3xsec2π2−5x=elimx→0sin2π2−3x−1sec2π2−5x =e−limx→0cos2π2−3xcos2π2−5x =e−limx→0sinπ2−π2−3xsinπ2−π2−5x2 =e−limx→0sin3πx2(2−3x)sin5πx2(2−5x)2 =e−352 =e−925

limx→0sin2π2−3xsec2π2−5x=

limx→0sin2π2−3xsec2π2−5x=elimx→0sin2π2−3x−1sec2π2−5x =e−limx→0cos2π2−3xcos2π2−5x =e−limx→0sinπ2−π2−3xsinπ2−π2−5x2 =e−limx→0sin3πx2(2−3x)sin5πx2(2−5x)2 =e−352 =e−925

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

$\lim_{x \to 0} {(\sin^{2} \frac{π}{2 - 3 x})}^{\sec^{2} \frac{π}{2 - 5 x}} =$

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

$e^{\frac{- 3}{5}}$

$e^{\frac{3}{5}}$

$e^{\frac{- 9}{25}}$

$e^{\frac{9}{25}}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{matrix} \lim_{x \to 0} {(\sin^{2} \frac{π}{2 - 3 x})}^{\sec^{2} \frac{π}{2 - 5 x}} = e^{\lim_{x \to 0} (\sin^{2} \frac{π}{2 - 3 x} - 1) \sec^{2} \frac{π}{2 - 5 x}} \\ = e^{- \lim_{x \to 0} \frac{\cos^{2} \frac{π}{2 - 3 x}}{\cos^{2} \frac{π}{2 - 5 x}}} \\ = e^{- \lim_{x \to 0} {(\frac{\sin (\frac{π}{2} - \frac{π}{2 - 3 x})}{\sin (\frac{π}{2} - \frac{π}{2 - 5 x})})}^{2}} \\ = e^{- \lim_{x \to 0} {(\frac{\sin (\frac{3 π x}{2 (2 - 3 x)})}{\sin (\frac{5 π x}{2 (2 - 5 x)})})}^{2}} \\ = e^{- {(\frac{3}{5})}^{2}} \\ = e^{\frac{- 9}{25}} \end{matrix}$