CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Evaluate limx0tan2xx3xsinx

Open in App
Solution

limx0tan2xx3xsinx=limx02sec22x13cosx=2131=12

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Property 1
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon