Answer
The limit does not exist (jump discontinuity)
Work Step by Step
$f(x)= \begin{cases}
1-x^2\quad \text{for }x\lt1\\
\dfrac{1}{x} \qquad\text{ for }x \geq1 \\
\end{cases}
$
We can test what kind of discontinuity we might have by first looking at the left and right hand limits.
$\lim\limits_{x \to 1^-} 1-x^2 = 1-(1^2) =0$
$\lim\limits_{x \to 1^+} \dfrac{1}{x} = \dfrac{1}{1} =1$
Because $\lim\limits_{x \to 1^-}f(x) \ne\lim\limits_{x \to 1^+}f(x)\quad\textbf{the limit does not exist.}$
So we can see from this that the function jumps from $0$ to $1$ at $x=1$.
So we have a $\textbf{jump discontinuity at x=1}$
We can graph this manually on paper.
For the first piece, $f(x)=1-x^2$, try plugging in values $0,-1$, and $-2$.
For the second piece, $f(x)= \dfrac{1}{x}$, try plugging in values $2, 3$, and $4$.
$\textbf{See graph}$
