Question

If $ f(x) = \begin{cases} \frac {Log x} {x-1} & \quad \text{if}\, x \neq 1\\ k & \quad \text{if} \,x=1\\ \end{cases} $ 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{cases} \frac{\log x}{x-1}, & \text{if $x \neq 1$ } \\ k, & \text{if $x=1$ } \end{cases}$ at x = 1
Since, the function is continuous at $x=1$. Then, $f(1)= \displaystyle \lim _{x \rightarrow 1} f(x)$
$k= \displaystyle\lim _{x \rightarrow 1} \frac{\log x}{x-1}\left(\frac{0}{0} \text { from }\right)$
Apply L'hospital rule,
$k= \displaystyle\lim _{x \rightarrow 1} \frac{1 / x}{1}=\frac{1 / 1}{1}$
$\Rightarrow k=1$