Answer
a) No value of $g(1)$;
b) $g(1)=-\dfrac{\pi}{2}$
Work Step by Step
We are given the function:
$g(x)=\tan^{-1}\left(\dfrac{1}{x-1}\right)$, $x\not=1$
a) Graph the function:
The function has a jump discontinuity at $x=0$, therefore we cannot make it continuous no matter which value $g(1)$ would take.
b) In order to make the function left-continuous in $x=1$, we should have:
$g(1)=\displaystyle\lim_x\rightarrow 1^{-} \tan^{-1}\left(\dfrac{1}{x-1}\right)=-\dfrac{\pi}{2}$
Therefore we must have:\\
$g(1)=-\dfrac{\pi}{2}$
