Question

If $f(x)=\{\begin{array}{ll}\frac{Logx}{x-1}& \text{if}x\ne 1\\ k& \text{if}x=1\end{array}$ is continuous at $x=1$, then the value of $k$ is

• A. 0
• B. -1
• C. 1
• D. e

Answer 1

Answer: (C)

$f(x)=\{\begin{array}{ll}\frac{\log x}{x-1},& \text{if x=1 }\\ k,& \text{if x=1 }\end{array}$ at x = 1
Since, the function is continuous at $x=1$. Then, $f(1)=\underset{x\to 1}{\lim }f(x)$
$k=\underset{x\to 1}{\lim }\frac{\log x}{x-1}\left(\frac{0}{0}\text{ from }\right)$
Apply L'hospital rule,
$k=\underset{x\to 1}{\lim }\frac{1/x}{1}=\frac{1/1}{1}$
$\Rightarrow k=1$