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

Evaluate the limit:
limx0cosec xcotxx

Open in App
Solution

We have,

limx0cosec xcotxx

=limx01sinxcosxsinxx

=limx01cosxxsinx

=limx02sin2x2xsinx
[1cosx=2sin2x2]

Dividing numerator and denominator by x2, we get

limx02×12×sinx2x2×12×sinx2x2sinxx

=2×12×1×12×11
[limx0sinxx=1]

=12

limx0cosecxcotxx=12

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon