JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\sin (\sin x)-\sin x}{a x^3+b x^5+c}=-\frac{1}{12}

, then

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Correct option: A

a=2, b \in R, c=0

Option A (

a=2, b \in R, c=0

) is correct because the limit exists and equals -1/12 only when a=2 and c=0, with b being any real number.

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