Question

It is given that ${\lim }_{x\to 0}\frac{e^{ax}-bx-1}{x^2}=2$ .Then the value of | a | + | b | is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

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