Question

Let $f\left(x\right)=\frac{|x^3-6x^2+11x-6|}{x^3-6x^2+11x-6}$, then the set of points ${}^{\prime }{a}^{\prime }$ where $\underset{x\to 0}{\lim }f(x)$ does not exist, are

• A. $1$, $2$, $3$
• B. $1$, $2$, $0$
• C. $-1$, $1$
• D. $0$, $1$

Answer 1

Answer: (A)

We have,
$f\left(x\right)=\left(\frac{|x-1|}{x-1}\right)\left(\frac{|x-2|}{x-2}\right)\left(\frac{|x-3|}{x-3}\right)$

$=\{\begin{array}{l}-1,x<1\\ 1,1<x<2\\ -1,2<x<3\\ 1,x>3\end{array}$
Therefore, the limits exists at all points except at $x=1$, $2$, $3$.
$\underset{x\to 1^{-}}{\lim }f(x)=-1$ and $\underset{x\to 1^{+}}{\lim }f(x)=1$
Since, $\underset{x\to 1^{-}}{\lim }f(x)\ne$ $\underset{x\to 1^{+}}{\lim }f(x)$
So, $\underset{x\to 1}{\lim }f(x)$ does not exist.
Similarly, $\underset{x\to a}{\lim }f(x)$ does not exist when $a=2$, $3$.