f(x) = begin{cases} frac{e^{alpha x} - e^{x} | vedclass

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(D) For the function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x 	o 0$ must equal $f(0)$.$\lim_{x 	o 0} \frac{e^{\alpha x} - e^{x}

Vedclass · Accepted Answer

$2$

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(D) For the function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x 	o 0$ must equal $f(0)$.$\lim_{x 	o 0} \frac{e^{\alpha x} - e^{x} - x}{x^{2}} = \frac{3}{2}$.Since the limit is in the $\frac{0}{0}$ form,we apply $L$'Hospital's rule:$\lim_{x 	o 0} \frac{\alpha e^{\alpha x} - e^{x} - 1}{2x} = \frac{3}{2}$.For the limit to exist,the numerator must also be $0$ at $x = 0$:$\alpha e^{0} - e^{0} - 1 = 0 \implies \alpha - 1 - 1 = 0 \implies \alpha = 2$.Checking with $\alpha = 2$ using $L$'Hospital's rule again:$\lim_{x 	o 0} \frac{2e^{2x} - e^{x} - 1}{2x} = \lim_{x 	o 0} \frac{4e^{2x} - e^{x}}{2} = \frac{4 - 1}{2} = \frac{3}{2}$.Thus,the value of $\alpha$ is $2$.

$f(x) = \begin{cases} \frac{e^{\alpha x} - e^{x} - x}{x^{2}}, & x \neq 0 \\ \frac{3}{2}, & x = 0 \end{cases}$ Find the value of $\alpha$ for which the function $f$ is continuous.

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