MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\alpha \;x}} - {e^{\beta \;x}}}}{x} = $
  • A
    $\alpha + \beta $
  • B
    $\frac{1}{\alpha } + \beta $
  • C
    ${\alpha ^2} - {\beta ^2}$
  • $\alpha - \beta $

Answer

Correct option: D.
$\alpha - \beta $
d
(d) $\mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\alpha x}} - {e^{\beta x}}}}{x} = \mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\alpha x}} - 1 - {e^{\beta x}} + 1}}{x}$

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\alpha x}} - 1}}{{x}} - \mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\beta x}} - 1}}{{ x}}$

$ = \alpha \mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\alpha x}} - 1}}{{\alpha x}} - \beta \mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\beta x}} - 1}}{{\beta x}}$

$ = \alpha .1 - \beta .1 = \alpha - \beta .$

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