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JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1
If lim⁡x→02ax+(a−1)sin⁡xtan⁡3x=l\lim \limits_{{x \rightarrow 0}} \frac{2 a x+(a-1) \sin x}{\tan ^3 x}=l, then a+la+l is equal to
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Correct option: D
49\frac{4}{9}
Option D (4/9) is correct because by applying L'Hopital's rule and standard limits, we find a=13a = \frac{{1}}{{3}} and l=19l = \frac{{1}}{{9}}, thus a+l=49a + l = \frac{{4}}{{9}}.

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