Question

The limiting value of $f\left(x\right)$ as $x$ approaches $1$, where $f\left(x\right)=\{\begin{array}{cc}4-x,& x<1\\ 4x-x^2,& x>1\end{array}$ is

• A. $0$
• B. $4$
• C. $3$
• D. does not exist

Answer 1

The correct option is C $3$

Given : $f\left(x\right)=\{\begin{array}{cc}4-x,& x<1\\ 4x-x^2,& x>1\end{array}$

For $x<1,f\left(x\right)=4-x$
$L.H.L.=\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(4-x)=4-1=3$

For $x>1,f\left(x\right)=4x-x^2$
$R.H.L.=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(4x-x^2)=4\left(1\right)-1^2=3$

Clearly, $L.H.L.=R.H.L.=3$
So, limiting value of $f\left(x\right)$ as $x$ approaches to $1$ is $3.$