Question

Evaluate $\underset{x\to 0}{\lim }\frac{a^x-b^x}{e^x-1}=$?

• A. ${\log }_e\left(\frac{a}{b}\right)$
• B. ${\log }_e\left(\frac{b}{a}\right)$
• C. ${\log }_e\left(ab\right)$
• D. ${\log }_e\left(a+b\right)$

Answer 1

The correct option is A

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

Explanation for the correct answer:

Simplifying the equation to determinate form and applying the limits:

Given, $\underset{x\to 0}{\lim }\frac{a^x-b^x}{e^x-1}$

Calculating the limit:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{a^x-b^x}{e^x-1} \\ =\underset{x\to 0}{\lim }\frac{a^x-1-b^x+1}{e^x-1}\mathbf{[}\mathbf{adding}\mathbf{}\mathbf{1}\mathbf{-}\mathbf{1}\mathbf{}\mathbf{in}\mathbf{}\mathbf{the}\mathbf{}\mathbf{numerator}\mathbf{]} \\ =\underset{x\to 0}{\lim }\frac{(a^x-1)-(b^x+1)}{e^x-1} \\ =\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{(a^x-1)-(b^x+1)}{x}}}{{\displaystyle \frac{e^x-1}{x}}}\mathbf{[}\mathbf{Dividing}\mathbf{}\mathbf{numerator}\mathbf{}\mathbf{and}\mathbf{}\mathbf{denominator}\mathbf{}\mathbf{by}\mathbf{}\mathbf{x}\mathbf{]} \\ =\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{a^x-1}{x}}-{\displaystyle \frac{b^x-1}{x}}}{{\displaystyle \frac{e^x-1}{x}}}\mathbf{[}\underset{\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{\because }\mathbf{lim}}\frac{{\mathbf{a}}^{\mathbf{x}}\mathbf{-}\mathbf{1}}{\mathbf{x}}\mathbf{=}\mathbf{lna}\mathbf{]}\mathbf{}\mathbf{} \end{gathered}$$

Applying the limits:

$$\begin{gathered} =\frac{\ln a-\ln b}{1} \\ =\ln \left(\frac{a}{b}\right)\left[\because \mathrm{lnx}-\mathrm{lny}=\ln \left(\frac{x}{y}\right)\right] \\ =\log \left(\frac{a}{b}\right) \end{gathered}$$

Therefore, the correct answer is option (A).