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Question

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

$\frac{14}{5}$
$\frac{3}{2}$
$\frac{5}{2}$
$\frac{7}{2}$

Solution

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