mathop {lim }limits_{x to 0} frac{{{a^x} - {b | vedclass

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(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}}$.Method $1$: Using standard limits.$\mathop {

Vedclass · Accepted Answer

$\log \left( {\frac{a}{b}} \right)$

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(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}}$.Method $1$: Using standard limits.$\mathop {\lim }\limits_{x 	o 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} = \mathop {\lim }\limits_{x 	o 0} \left( \frac{\frac{a^x - 1}{x} - \frac{b^x - 1}{x}}{\frac{e^x - 1}{x}} ight)$.Using the standard limit $\mathop {\lim }\limits_{x 	o 0} \frac{k^x - 1}{x} = \ln k$,we get:$= \frac{\ln a - \ln b}{\ln e} = \ln \left( \frac{a}{b} ight)$.Method $2$: Using $L-Hospital's$ rule.Since the form is $\frac{0}{0}$,differentiate numerator and denominator with respect to $x$:$= \mathop {\lim }\limits_{x 	o 0} \frac{a^x \ln a - b^x \ln b}{e^x} = \frac{a^0 \ln a - b^0 \ln b}{e^0} = \ln a - \ln b = \ln \left( \frac{a}{b} ight)$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} = $

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