Let $\beta=\lim_{x \rightarrow 0} \frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)}$ for some $\alpha \in \mathbb{R}$. Then the value of $\alpha+\beta$ is:

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Limits

Let

β = lim_{x \to 0} \frac{α x - ( e ^{3 x} - 1 )}{α x ( e ^{3 x} - 1 )}

for some

α \in R

. Then the value of

α + β

is:

Answer: (C)

Expand

e^{3 x} = 1 + 3 x + \frac{9 x ^{2}}{2} + \dots

β = lim_{x \to 0} \frac{α x - 3 x - \frac{9 x ^{2}}{2} - \dots}{α x \cdot 3 x \cdot ( 1 + \frac{3 x}{2} + \dots )}

For the limit to exist, the numerator's coefficient of

x

must be zero:

α - 3 = 0

, so

α = 3

.

With

α = 3

β = lim_{x \to 0} \frac{- \frac{9 x ^{2}}{2}}{9 x ^{2}} = - \frac{1}{2}

Therefore,

α + β = 3 - \frac{1}{2} = \frac{5}{2}

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option

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