Question

If $f\left(x\right)=\{\begin{array}{cc}7-4x,& x<1\\ x^2+2,& x\ge 1\end{array}$, then value of $\underset{x\to 1}{lim}f\left(x\right)$ is

• A. $3$
• B. $-3$
• C. it does not exist
• D. $1$

Answer 1

The correct option is A $3$

$L.H.L.=\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(7-4x)=3(\because x\to 1^{-}\text{implies that}x<1)$

$R.H.L.=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(x^2+2)=3(\because x\to 1^{+}\text{implies that}x>1)$
Clearly,
$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=3$
Hence, $\underset{x\to 1}{lim}f\left(x\right)=3$