Skip to main content
Back to LIMITS

JEE MATHEMATICS MCQ: LIMITS

1 Easy
1 of 1
Application +4 | -1
lim⁡x→π/62sin⁡2x+sin⁡x−12sin⁡2x−3sin⁡x+1=\lim \limits_{{x \rightarrow \pi / 6}} \frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}=
View answer and explanation
Correct option: B
-3
Option B (-3) is correct because after factoring and simplifying the limit expression, substituting x = pi/6 yields -3.

More Questions from LIMITS

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.