Question

The value of $a$ so that
$\underset{x\to 0}{lim}\frac{1}{x^2}(e^{ax}-e^x-x)=\frac{3}{2}$ is

• A. $1$
• B. $0$
• C. $4$
• D. $2$

Answer 1

The correct option is D $2$

Since the numerator tends to 0 as $x\to 0$ so $\underset{x\to 0}{lim}\frac{1}{x^2}(e^{ax}-e^x-x)=\frac{1}{2}\underset{x\to 0}{lim}\frac{(\alpha e^{ax}-e^x-1)}{x}$.

For the last limit to exist we must have $\underset{x\to 0}{lim}(\alpha e^{ax}-e^x-1)=0.\therefore \alpha -1-1=0\Rightarrow \alpha =2$.

For $\alpha =2$, the last limit is equal to $\frac{1}{2}\underset{x\to 0}{lim}\frac{(2e^{2x}-e^x-1)}{x}$
$=\frac{1}{2}\underset{x\to 0}{lim}(4e^{2x}-e^x)=\frac{3}{2}$