JEE MATHEMATICS MCQ: LIMITS

1 Easy

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Application +4 | -1

f(3)=6

f^{\prime}(3)=2

, then

\operatorname{Limit}_{x \rightarrow 3} \frac{x f(3)-3 f(x)}{x-3}

is given by :

View answer and explanation

Correct option: C

Option C (0) is correct because when applying L'Hopital's rule and using the given values of f(3) and f'(3), the limit evaluates to zero.

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\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n}