Question

If $g\left(x\right)=\{\begin{array}{cc}-x^2& \text{if}x<-1\\ 3& \text{if}x=-1\\ 2-x& \text{if}-1<x\le 1\\ x^2& \text{if}x>1\end{array}$ then

• A. $g$ is not continuous at $x=-1$
• B. $\underset{x\to 1}{lim}g\left(x\right)<\underset{x\to 0}{lim}\frac{e^x-1}{\sin x}$
• C. $g$ is not continuous at $x=1$
• D. $g$ is continuous at $x=-1$

Answer 1

The correct option is A $g$ is not continuous at $x=-1$

$g\left(x\right)=\{\begin{array}{cc}-x^2& ifx<-1\\ 3& ifx=-1\\ 2-x& if-1<x\le 1\\ x^2& ifx>1\end{array}{lim}_{x\to -1^{-}}g\left(x\right)={lim}_{x\to -1^{-}}-x^2=-1{lim}_{x\to -1^{+}}g\left(x\right)={lim}_{x\to -1^{+}}2-x=3{lim}_{x\to 1^{-}}g\left(x\right)={lim}_{x\to 1^{-}}2-x=2-1=1{lim}_{x\to 1^{+}}g\left(x\right)={lim}_{x\to 1^{+}}{x}^2=1$
Hence $g\left(x\right)$ is not continuous at $x=-1$ and continuous at $x=1$