Question

If $f\left(x\right)=\{\begin{array}{cc}x+1,& x>0\\ 2-x,& x\le 0\end{array}$ and $g\left(x\right)=\{\begin{array}{cc}x+3,& x<1\\ x^2-2x-2,& 1\le x<2\\ x-5,& x\ge 2\end{array}$, then the value of $\underset{x\to 0}{\text{lim}}g\left(f\right(x\left)\right)$ is

• A. $2$
• B. $1$
• C. $-3$
• D. It does not exist

Answer 1

The correct option is C $-3$

As $x\to 0^{-},f\left(x\right)\to f\left(0^{-}\right)=2^{+}.$

$\Rightarrow \underset{x\to 0^{-}}{\text{lim}}g\left(f\right(x\left)\right)=g\left(2^{+}\right)=2-5=-3$

Also, as $x\to 0^{+},f\left(x\right)\to f\left(0^{+}\right)=1^{+}$

$\Rightarrow \underset{x\to 0^{+}}{\text{lim}}g\left(f\right(x\left)\right)=g\left(1^{+}\right)=1^2-2\cdot 1-2=-3$

$\therefore \underset{x\to 0^{-}}{\text{lim}}g\left(f\right(x\left)\right)=\underset{x\to 0^{+}}{\text{lim}}g\left(f\right(x\left)\right)=\underset{x\to 0}{\text{lim}}g\left(f\right(x\left)\right)=-3$