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

The value of limx0cos(sinx)cosxx4=

A
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
16
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
16
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 16
limx 0cos(sinx)cosxx4
Rewriting the difference in the numerator as a product
cosθcosϕ2sinθ+ϕ2sinϕθ2
limx 02sinx+sinx2sinxsinx2x4
2limx 0[1x4×sin(x+sinx2)×sin(xsinx2)]
2limx 0[1x4sin(x+sinx2)(x+sinx2)×x+sinx2×sin(xsinx2)(x+sinx2)×xsinx2
2limx 0sin(x+sinx2)(x+sinx2)×limx 0sin(xsinx2)(xsinx2)×limx 0[1x4×(x+sinxxsinx)4]
2×1×1limx 0[1x4×x+sinx2×xsinx2]
12limx 0[x2sin2xx4]
Replacing \sin x by Maclarian series:-
sinx=xx33!+x55!
sinx=x22x43!+p(x)
We do not need exact expression of p(x)
12limx 0[x2sin2xx4]
12limx 0[x2x22x43!+p(x)x4]
12limx 0[2x43!p(x)x4]
p(x) contains x\,power >4
p(x)0
12limx 0[23!0]=16

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