Question

The value of ${lim}_{x\to a}\frac{\log (x-a)}{\log (e^x-e^a)}$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{1}{a}$

Answer 1

Answer: (D)

$\frac{1}{a}$

${lim}_{x\to a}\frac{\log (x-e)}{\log (e^x-e^x)}$

Applying L'Hospital Rule,

${lim}_{x\to a}\frac{\frac{d}{x}\left(\log \right(x-o\left)\right)}{\frac{d}{dx}(\log e^x-e^a)}$

$={lim}_{x\to a}\frac{\frac{1}{x-a}}{\frac{e^x}{e^x-e^a}}={lim}_{x\to a}\frac{e^x-e^a}{e^x(x-a)}$

Applying L'Hospital Rule again

$={lim}_{x\to a}\frac{e^x}{e^x+x{e}^x-e^x}$

$={lim}_{x\to a}\frac{e^x}{x{e}^x}$

$={lim}_{x\to a}\frac{1}{x}$

$=\frac{1}{a}$