We have, lim_ x→ 0a^x b^xe^x 1 This is the 00 So, apply L Hospital rule lim_ x→ 0a^x ln a b^xln be^x 0 lim_ x→ 0a^x ln a b^xln be^x =a^0 ...

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Algebra of Limits

limx → 0ax-bx...

Question

$l i m_{x \to 0} \frac{a^{x} - b^{x}}{e^{x} - 1}$

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Solution

We have,
${lim}_{x \to 0} \frac{a^{x} - b^{x}}{e^{x} - 1}$

This is the $\frac{0}{0}$
So, apply L-Hospital rule
${lim}_{x \to 0} \frac{a^{x} l n a - b^{x} ln b}{e^{x} - 0}$
${lim}_{x \to 0} \frac{a^{x} l n a - b^{x} ln b}{e^{x}}$
$= \frac{a^{0} l n a - b^{0} ln b}{e^{0}}$
$= \frac{l n a - l n b}{1}$
$= l n a - l n b$
$= l n \frac{a}{b}$

Hence, this is the answer.

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