f x = e α x − e x − x x 2 , x ≠ 0 3 2 , x = 0 Find the value of α for which the…

For this function to be continuous

$\underset{x\to 0}{\lim }\frac{e^{\alpha x}-e^x-x}{x^2}=\frac{3}{2}$,     $\left(\text{it is }\frac{0}{0} \text{form}\right)$

Now, applying L'Hospital rule,

$\underset{x\to 0}{\lim }\frac{\alpha e^{\alpha x}-e^x-1}{2x}=\frac{\alpha e^{\alpha \left(0\right)}-e^{\left(0\right)}-1}{2x}$

Now, as denominator gives the value $0$.

In order for the limit to exist, the value of numerator,

$\alpha e^{\alpha \left(0\right)}-e^{\left(0\right)}-1=0\Rightarrow \alpha =2$