Question

Let $\beta =\underset{x\to 0}{\lim }\frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)}$ 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)

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