Question

For the figure given below, which of the following is/are true

• A. $\underset{x\to 3}{lim}f\left(x\right)=2$
• B. $\underset{x\to 4}{lim}f\left(x\right)=4$
• C. $\underset{x\to 5}{lim}f\left(x\right)=4$
• D. $\underset{x\to 6}{lim}f\left(x\right)=0$

Answer 1

Answer: (B), (C)

$\underset{x\to 5}{lim}f\left(x\right)=4$

From the above graph
For $x\to 3$
$L.H.L.=\underset{x\to 3^{-}}{lim}f\left(x\right)=2R.H.L.=\underset{x\to 3^{+}}{lim}f\left(x\right)=3\because L.H.L.\ne R.H.L.$
$\therefore \underset{x\to 3}{lim}f\left(x\right)$ does not exist

For $x\to 4$
$L.H.L.=\underset{x\to 4^{-}}{lim}f\left(x\right)=4R.H.L.=\underset{x\to 4^{+}}{lim}f\left(x\right)=4\because L.H.L.=R.H.L.$
$\therefore \underset{x\to 4}{lim}f\left(x\right)=4$

For $x\to 5$
$L.H.L.=\underset{x\to 5^{-}}{lim}f\left(x\right)=4R.H.L.=\underset{x\to 5^{+}}{lim}f\left(x\right)=4\because L.H.L.=R.H.L.$
$\therefore \underset{x\to 5}{lim}f\left(x\right)=4$

For $x\to 6$
$L.H.L.=\underset{x\to 6^{-}}{lim}f\left(x\right)=-\infty R.H.L.=\underset{x\to 6^{+}}{lim}f\left(x\right)=0\because L.H.L\ne R.H.L.$
$\therefore \underset{x\to 6}{lim}f\left(x\right)$ does not exist