JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

\lim\limits_{x \to \frac{\pi }{6}} \,\,\frac{{2{{\sin }^2}x + \sin x - 1}}{{2{{\sin }^2}x - 3\sin x + 1}}

is equal to

View answer and explanation

Correct option: B

- 3

Option B (-3) is correct because the limit is evaluated using L'Hopital's rule or by factoring and simplifying the expression.

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