wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

limxπ4f(x)f(π4)xπ4, where f(x)=sin2x

Open in App
Solution

Solution -
limxπ4sin2x1xπ4
Apply L-Hospital Rule
limxπ42cosx
=2×0
=0

1092185_1174543_ans_09b8c80489034f9f819582c400240afe.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Introduction to Limits
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon