Question

Evaluate:$\underset{x\to 1}{\lim }{\left(\log ex\right)}^{\frac{1}{\log x}}$

• A. $e^{-1}$
• B. $e$
• C. $2$
• D. $0$

Answer 1

The correct option is B

$e$

Find the value of $\underset{x\to 1}{\lim }{\left(\log ex\right)}^{\frac{1}{\log x}}$

Consider the given Equation as: $y=\underset{x\to 1}{\lim }{\left(\log ex\right)}^{\frac{1}{\log x}}$

Then,

$$\begin{gathered} y=\underset{x\to 1}{\lim }{\left(\log e+\log x\right)}^{\frac{1}{\log x}} \\ y=\underset{x\to 1}{\lim }{\left(1+\log x\right)}^{\frac{1}{\log x}}[\because \log \left(e\right)=1] \end{gathered}$$

Taking $\log$on both sides,

$\log y=\underset{x\to 1}{\lim }\log {\left(1+\log x\right)}^{\frac{1}{\log x}}$

We know that

$\log a^b=b\log a$, therefore

$$\begin{gathered} \log y=\underset{x\to 1}{\lim }\left(\frac{1}{\log x}\log \left(1+\log x\right)\right) \\ \text{Putting}x\to 1 \\ \log y=\left(\frac{\log \left(1+\log 1\right)}{\log 1}\right) \\ \log y=\left(\frac{\log \left(1+0\right)}{0}\right)[\because \log \left(1\right)=0] \\ \log y=\frac{0}{0}\left[\text{Indeterminate form}\right] \end{gathered}$$

Using the L' Hospital rule

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

$$\begin{gathered} \log y=\underset{x\to 1}{\lim }\frac{{\displaystyle \frac{1}{1+\log x}}.{\displaystyle \frac{1}{x}}}{{\displaystyle \frac{1}{x}}} \\ \log y=\underset{x\to 1}{\lim }\frac{1}{1+\log x} \\ \log y=\frac{1}{1+\log \left(1\right)} \\ \log y=\frac{1}{1+0} \\ \log y=1 \\ y=e \end{gathered}$$

Hence, the correct answer is Option B