mathop {lim }limits_{x to 0} frac{{{e^{alpha | vedclass

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Question

(D) We use the standard limit formula $\mathop {\lim }\limits_{u 	o 0} \frac{e^u - 1}{u} = 1$. $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^{\alpha x}

Vedclass · Accepted Answer

$\alpha - \beta $

Vedclass · Answer

(D) We use the standard limit formula $\mathop {\lim }\limits_{u 	o 0} \frac{e^u - 1}{u} = 1$. $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^{\alpha x}} - {e^{\beta x}}}}{x} = \mathop {\lim }\limits_{x 	o 0} \frac{{{e^{\alpha x}} - 1 - ({e^{\beta x}} - 1)}}{x}$ $= \mathop {\lim }\limits_{x 	o 0} \left( \frac{{{e^{\alpha x}} - 1}}{x} - \frac{{{e^{\beta x}} - 1}}{x} ight)$ $= \mathop {\lim }\limits_{x 	o 0} \left( \alpha \cdot \frac{{{e^{\alpha x}} - 1}}{{\alpha x}} - \beta \cdot \frac{{{e^{\beta x}} - 1}}{{\beta x}} ight)$ $= \alpha(1) - \beta(1) = \alpha - \beta$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\alpha x}} - {e^{\beta x}}}}{x} = $

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