Question

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

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

Answer 1

Answer: (A)

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

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