wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The value of limx06sinx+3sin2x4sin3xx2sinx is

Open in App
Solution

limx06sinx+3sin2x4sin3xx2sinx=limx06sinx+6sinxcosx4(3sinx4sin3x)x2sinx=limx06+6cosx12+16sin2xx2=limx06cosx6+16sin2xx2=16limx0sin2xx26limx01cosxx2=163=13

Alternate Solution:

L=limx06sinx+3sin2x4sin3xx2sinx=limx06(xx33!+...)+3(2x(2x)33!+...)4(3x(3x)33!+...)x2(xx33!+...)=limx06x6x33!+6x24x33!12x+108x33!+.....x3(1x23!+...)=624+1083!=13




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