Question

If $g\left(x\right)=\{\begin{array}{cc}{x}^2+5& \text{if}x<2\\ 10& \text{if}x=2\\ (1+x^3)/(1-x)& \text{if}x>2\end{array}$, then

• A. $\underset{x\to 2+}{lim}g\left(x\right)=-9$
• B. $\underset{x\to 2-}{lim}g\left(x\right)=9$
• C. $\underset{x\to 2+}{lim}g\left(x\right)=-3$
• D. $g$ is continuous at $x=2$

Answer 1

Answer: (A), (B)

The correct options are
A $\underset{x\to 2+}{lim}g\left(x\right)=-9$
B $\underset{x\to 2-}{lim}g\left(x\right)=9$

$g\left(x\right)=\{\begin{array}{cc}{x}^2+5& ifx<2\\ 10& ifx=2\\ (1+x^3)/(1-x)& ifx>2\end{array}\underset{x\to 2+}{lim}g\left(x\right)=\underset{x\to 2+}{lim}\frac{1+x^3}{1-x}=-9\underset{x\to 2-}{lim}g\left(x\right)=\underset{x\to 2}{lim}4+5=9$

L.H.L $\ne$ R.H.L

$\therefore g\left(x\right)$ is not continuous at $x=2$

Only option A and B are correct.