## Calculus (3rd Edition)

a) No value of $g(1)$; b) $g(1)=-\dfrac{\pi}{2}$
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:\\ $g(1)=-\dfrac{\pi}{2}$