wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

Solve it:
limxπ22sinx1(π2x)

Open in App
Solution

Solution -
limxπ22sinx1(π2x)

limxπ21sinxπ2x

Apply L-Hospital rule

limxπ2cosx21sinx
limxπ2cos1+sinx21sinx1+sinx

limxπ21+sinx2

=22

=12

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Fundamental Theorem of Calculus
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon