Question

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

• A. $3$
• B. $2$
• C. Does not exist
• D. $1$

Answer 1

The correct option is A $3$

$L.H.L.=\underset{x\to 1^{-}}{lim}f\left(x\right)$
$=\underset{x\to 1^{-}}{lim}(7-4x)$
$=\underset{h\to 0}{lim}(7-4(1-h\left)\right)$
$=\underset{h\to 0}{lim}(3+4h)=3$

$R.H.L.=\underset{x\to 1^{+}}{lim}f\left(x\right)$
$=\underset{x\to 1^{+}}{lim}(x^2+2)$
$=\underset{h\to 0}{lim}\left(\right(1+h{)}^2+2)$
$=3$

Clearly,
$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=3$
Hence, $\underset{x\to 1}{lim}f\left(x\right)=3$