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JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 71

If f(x)f(x) is continuous and f(92)=29f\left(\frac92\right)=\frac29, then lim⁡x→0f(1−cos⁡3xx2)\displaystyle\lim_{x\to0}f\left(\frac{1-\cos3x}{x^2}\right) is equal to :

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Official answer: B

1−cos⁡3xx2→92\dfrac{1-\cos3x}{x^2}\to\dfrac92. By continuity, the limit is f(92)=29f\left(\frac92\right)=\frac29.

Topics:LimitsContinuity
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