Question

Consider $f\left(x\right)=sgn\left(sinx\right)+\left\{x\right\};2\le x\le 4$ and $g\left(x\right)=-2+\mid x-3\mid ;$ where $\{.\}$ denotes fractional part function and $sgn$ denotes signum function. Then $\underset{x\to 3}{lim}gof\left(x\right)$ is equal to

• A. -1
• B. 0
• C. 1
• D. does not exist

Answer 1

The correct option is D does not exist

First write $f\left(x\right)$ and $g\left(x\right)$ in different intervals as per the definition.

$f\left(x\right)=\{\begin{array}{c}x-1,2\le x<3\\ x-2,3\le x<\pi \\ x-3,x=\pi \\ x-4,\pi <x\le 4\end{array}$

$g\left(x\right)=\{\begin{array}{c}1-x,x<3\\ x-5,x\ge 3\end{array}$

At $x=3$, $RHL=gof\left(3^{+}\right)=\underset{h\to 0}{lim}g\left(f\right(3+h\left)\right)=\underset{h\to 0}{lim}g(3+h-2)=\underset{h\to 0}{lim}g(1+h)$

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

$LHL=gof\left(3^{-}\right)=\underset{h\to 0}{lim}g\left(f\right(3-h\left)\right)=\underset{h\to 0}{lim}g(3-h-1)=\underset{h\to 0}{lim}g(2-h)$

$=\underset{h\to 0}{lim}1-(2-h)=\underset{h\to 0}{lim}-1+h=-1$
Since $LHL\ne RHL$, limit does not exist.