Answer
$-6,0$
Work Step by Step
$(f o g) (x)=\dfrac{4}{1/x+2-1}=\dfrac{4(x+2)}{-1-x}$
Now, $\lim_\limits{x \to 1} (f o g) (x)=\dfrac{4(1+2)}{-1-1}=-6$
and $( g o f) (x)=\dfrac{1}{\dfrac{4}{x-1}-1}=\dfrac{x-1}{4+2(x-1)}$
Now, $\lim_\limits{x \to 1} ( g o f) (x)=\dfrac{1-1}{4+2(1-1)}=0$
Answers: $-6,0$
