Question

Let $f\left(x\right)=\{\begin{array}{cc}x+1,& x>0\\ 2-x,& x\le 0\end{array}$ and $g\left(x\right)=\{\begin{array}{cc}x+3,& x<1\\ x^2-2x-2,& 1\le x<2\\ x-5,& x\ge 2\end{array}$.
Find $\underset{x\to 0}{lim}g\left(f\right(x\left)\right)$.

• A. $-3$
• B. $-2$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$-3$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)$

$=\underset{h\to 0}{lim}2-(-h)$

$=\underset{h\to 0}{lim}2+h$
Now,$\underset{x\to 0^{-}}{lim}g\left(f\right(x\left)\right)$ $=\underset{h\to 0}{lim}g(2+h)$

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

$=\underset{h\to 0}{lim}h+1$
Now,$\underset{x\to 0^{+}}{lim}g\left(f\right(x\left)\right)$

$=\underset{h\to 0}{lim}g(1+h)$

$=\underset{h\to 0}{lim}{(1+h)}^2-2(1+h)-2=-3$
Hence, LHL=RHL

$\Rightarrow \underset{x\to 0}{lim}g\left(f\right(x\left)\right)=-3$