Question

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

• A. $\frac{14}{5}$
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $\frac{7}{2}$

Answer 1

Answer: (C)

$\beta=\displaystyle\lim _{x \rightarrow 0} \frac{\alpha x-\left(e^{3 x}-1\right)}{\alpha x\left(e^{3 x}-1\right)} $
$ \beta=\displaystyle\lim _{x \rightarrow 0} \frac{1+\alpha x-\left[1+3 x+\frac{9 x^2}{2 !}+\ldots . .\right]}{(\alpha x) \frac{\left(e^{3 x}-1\right)}{3 x} 3 x} $
$\beta=\displaystyle\lim _{x \rightarrow 0} \frac{(\alpha x-3 x)-\frac{9 x^2}{2 !}-\ldots \ldots}{3 \alpha x^2}$
For existence of limit $\alpha-3=0$
$\alpha=3$
Limit $\beta=\frac{-3}{2 \alpha}$
$\beta=-\frac{1}{2}$
Now,
$\alpha+\beta=\frac{5}{2}$