Question

It is given that $\lim_{x\to0} \frac{e^{ax}-bx-1}{x^{2}} = 2 $ .Then the value of | a | + | b | is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

We have,
$\lim\limits_{x\to0} \frac{e^{ax} -bx -1}{x^{2}} = 2 $
Apply L, Hospital’s rule, we get
$ \lim \limits_{x\to 0} \frac{ae^{ax} -b}{2x}$
$\therefore a -b = 0$
Again apply L, Hospital’s rule, we get
$ \lim\limits_{x\to0}\frac{ a^{2}e^{ax}}{2} = 2 $
$ ⇒ a^{2} = 4 $
$⇒ a= \pm 2$
$ ∴ b = \pm 2$
$ | a| + \left|b\right| = 2 + 2 =4$