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JEE MATHEMATICS MCQ: LIMITS

1 Easy
1 of 1
Application +4 | -1
lim⁡x→02sin⁡23xx2\lim\limits_{x \to 0} \frac{{2{{\sin }^2}3x}}{{{x^2}}} is equal to
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Correct option: C
18
Option C (18) is correct because by using the limit lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{{\sin x}}{x} = 1, the given limit simplifies to 18.

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