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Question

Question: \[\lim_{x \rightarrow 0}\frac{e^{\alpha x} - e^{\beta x}}{x} =\]...

lim⁡x→0eαx−eβxx=\lim_{x \rightarrow 0}\frac{e^{\alpha x} - e^{\beta x}}{x} =

A

α+β\alpha + \beta

B

1α+β\frac{1}{\alpha} + \beta

C

α2−β2\alpha^{2} - \beta^{2}

D

α−β\alpha - \beta

Answer

α−β\alpha - \beta

Explanation

Solution

lim⁡x→0eαx−eβxx=lim⁡x→0(eαx−1)−(eβx−1)x\lim_{x \rightarrow 0}\frac{e^{\alpha x} - e^{\beta x}}{x} = \lim_{x \rightarrow 0}\frac{(e^{\alpha x} - 1) - (e^{\beta x} - 1)}{x}

=lim⁡x→0eαx−1x−lim⁡x→0eβx−1x\lim_{x \rightarrow 0}\frac{e^{\alpha x} - 1}{x} - \lim_{x \rightarrow 0}\frac{e^{\beta x} - 1}{x} = α−β\alpha - \beta.