JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

\lim\limits_{x \to 0} \frac{{\left( {{e^{kx}} - 1} \right)\sin kx}}{{{x^2}}} = 4

, then

k

is equal to

View answer and explanation

Correct option: C

\pm \,2

Option C (

\pm \,2

) is correct because solving the limit equation yields two possible values for k, both positive and negative 2.

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