Question

$f\left(x\right)=\{\begin{array}{cc}4x-3,& x<1\\ x^2& x\ge 1\end{array},$ then
$\underset{x\to 1}{lim}f\left(x\right)=$

• A. $1$
• B. $-1$
• C. $0$
• D. $0$

Answer 1

The correct option is A $1$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(4x-3)=1$
$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}{x}^2=1$
Since LHL = RHL
$\Rightarrow \underset{x\to 1}{lim}f\left(x\right)=1$