Question

$\underset{x\to a}{\lim }\left(\frac{\log \left(x-a\right)}{\log \left(e^x-e^a\right)}\right)=$

• A. $0$
• B. $1$
• C. $a$
• D. dose not exist

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Find the value of $\underset{x\to a}{\lim }\left(\frac{\log \left(x-a\right)}{\log \left(e^x-e^a\right)}\right)$

The given equation is,

$$\begin{gathered} I=\underset{x\to a}{\lim }\left(\frac{\log \left(x-a\right)}{\log \left(e^x-e^a\right)}\right) \\ I=\left(\frac{\log \left(a-a\right)}{\log \left(e^a-e^a\right)}\right) \\ I=\frac{\log \left(0\right)}{\log \left(0\right)} \\ I=\frac{-\infty }{-\infty } \end{gathered}$$

Using the L' Hospitality rule

$\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

$$\begin{gathered} I=\underset{x\to a}{\lim }\left(\frac{{\displaystyle \frac{1}{\left(x-a\right)}}.\left(1\right)}{{\displaystyle \frac{1}{\left(e^x-e^a\right)}}.e^x}\right) \\ I=\underset{x\to a}{\lim }\left(\frac{{\displaystyle e^x-e^a}}{{\displaystyle e^x\left(x-a\right)}}\right) \\ I=\left(\frac{{\displaystyle e^a-e^a}}{{\displaystyle e^a\left(a-a\right)}}\right) \\ I=\frac{0}{0}\left[\text{form}\right] \end{gathered}$$

Using the L' Hospitality rule

$$\begin{gathered} I=\underset{x\to a}{\lim }\left(\frac{{\displaystyle e^x-0}}{{\displaystyle e^x\left(x-a\right)+e^x\left(1-0\right)}}\right) \\ I=\left(\frac{{\displaystyle e^a-0}}{{\displaystyle e^a\left(a-a\right)+e^a\left(1\right)}}\right) \\ I=\left(\frac{{\displaystyle e^a}}{{\displaystyle 0+e^a}}\right) \\ I=1 \end{gathered}$$

Hence, the correct answer is option (B).