Question

Consider $f\left(x\right)=sgn\left(\sin x\right)+\left[x\right];2\le x\le 4$ and $g\left(x\right)=-2+|x-3|$; where $[.]$ denotes greatest integer function. Then $\underset{x\to 3}{lim}gof\left(x\right)$ equal to

Answer 1

Dividing the region in four parts

$\to [2,3);[3,\pi );[\pi ,4);\left\{4\right\}$

$f\left(x\right)$$=\{3,2\le x<33,3\le x<\pi 2.\pi \le x<43,x=4\}$

$g\left(x\right)$$=\{-x+1,2\le x<3x-5,3\le x\le 4\}$

$li{m}_{x\to 3^{+}}g\circ f\left(x\right)=g\left(3\right)=3-5=-2$

and $li{m}_{x\to 3^{-}}g\circ f\left(x\right)=g\left(3\right)=3-5=-2$

since $li{m}_{x\to 3^{-}}g\circ f\left(x\right)=$$li{m}_{x\to 3^{+}}g\circ f\left(x\right)=-2$

so $li{m}_{x\to 3}g\circ f\left(x\right)=g\left(3\right)=-2$