First slide

Introduction to limits

Question

$If L = lim_{x \to 0} \frac{e^{a x} - e^{x} - x}{x^{2}} is a finite real number, then$

Moderate

A

$a = 2, L = 0$
B

$a = 0, L = \frac{3}{2}$
C

$a = 2, L = \frac{3}{2}$
D

$a = 0, L = 2$

Solution

$\begin{array}{l} Given L = lim_{x \to 0} \frac{e^{a x} - e^{x} - x}{x^{2}} = finite (\frac{0}{0} form) \\ Using L'Hospitals rule \\ L = lim_{x \to 0} \frac{e^{a x} a - e^{x} - 1}{2 x} \dots \dots \dots \dots \dots . . . (1) \end{array}$
$\begin{array}{l} (\begin{matrix} ∵ \frac{a - 1 - 1}{2 (0)} = finite \\ \Rightarrow \frac{a - 2}{0} = finite \end{matrix}) \\ ∴ a - 2 = 0 \Rightarrow a = 2 \\ Substituting a = 2 in (1) \\ L = lim_{x \to 0} \frac{2 e^{2 x} - e^{x} - 1}{2 x} (\frac{0}{0} form) \end{array}$
$\begin{array}{l} = lim_{x \to 0} \frac{4 e^{2 x} - e^{x}}{2} \\ = \frac{4 - 1}{2} \\ = \frac{3}{2} \\ ∴ a = 2, L = \frac{3}{2} \end{array}$

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