Question

If $\underset{x\to 0}{lim}\frac{\log (x+a)-\log a}{x}+k\underset{x\to e}{lim}\frac{\log x-1}{x-e}=1$, then the value of $k$ is:

• A. $1-\frac{1}{a}$
• B. $e\left(1-a\right)$
• C. $e\left(1-\frac{1}{a}\right)$
• D. $e\left(1+a\right)$

Answer 1

The correct option is C

$e\left(1-\frac{1}{a}\right)$

Explanation for the correct option.

Step 1: Solve $\underset{x\to 0}{lim}\frac{\log (x+a)-\log a}{x}$

$$\begin{gathered} \underset{x\to 0}{lim}\frac{\log (x+a)-\log a}{x} \\ =\frac{\log (0+a)-\log a}{0} \\ =\frac{\log a-\log a}{0} \\ =\frac{0}{0}form \end{gathered}$$

Step 2: Apply L's Hospital Rule

As, $\underset{x\to 0}{lim}\frac{\log (x+a)-\log a}{x}$ is in $\frac{0}{0}$ form, so we will apply L's Hospital Rule, and we get

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

Step 3: Solve $\underset{x\to e}{lim}\frac{\log x-1}{x-e}$

$$\begin{gathered} \underset{x\to e}{lim}\frac{\log x-1}{x-e} \\ =\frac{\log e-1}{e-e} \\ =\frac{1-1}{0} \\ =\frac{0}{0}form \end{gathered}$$

Step 4: Apply L's Hospital Rule

As, $\underset{x\to e}{lim}\frac{\log x-1}{x-e}$ is in $\frac{0}{0}$ form, so we will apply L's Hospital Rule, and we get

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

Step 5: Find the value of $k$.

$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{\log (x+a)-\log a}{x}+k\underset{x\to e}{lim}\frac{\log x-1}{x-e}& =& 1\\ \Rightarrow \frac{1}{a}+k\frac{1}{e}& =& 1\\ \Rightarrow \frac{k}{e}& =& 1-\frac{1}{a}\\ \Rightarrow k& =& e\left(1-\frac{1}{a}\right)\end{array}$

Hence, option C is correct.