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

The value of limxπ2sinxtanx is


A

1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

e

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

None of these.

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

1


Explanation of the correct option.

Step 1: Put the values of limit.

Given :limxπ2sinxtanx

Let L=limxπ2sinxtanx

Taking ln both side,

lnL=limxπ2lnsinxtanxlnL=limxπ2tanxlnsinxlnL=limxπ2lnsinxcotxlnL=00

Step 2: Apply L. Hospital's rule.

Since, it is 00 form, apply L. Hospital's rule,

lnL=limxπ21sinx.cosx-cosec2x

lnL=limxπ2-cosx×sin2xsinxlnL=0L=e0L=1

Therefore, the value of limxπ2sinxtanx is 1.

Hence, option A is the correct option, i.e. 1.


flag
Suggest Corrections
thumbs-up
5
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Basic Theorems in Differentiation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon