Answer
Removable discontinuity at $x=0$
Work Step by Step
$f(x) =
\begin{cases}
\cos(x)\quad\text{for x<0}\\
0\quad\quad\quad\text{for x=0}\\
1-x^2\quad\text{for x>0}
\end{cases}$
Let's start by evaluating the left and right hand limits.
$\lim\limits_{x \to 0^-}\cos(x) = \cos(0)= 1$
$\lim\limits_{x \to 0^+} = 1-x^2 = 1-(0)^2 = 1$
Since $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+}f(x)$ we can say that
$\lim\limits_{x \to 0}f(x)=1$.
But $f(0) = 0$.
Since $\textbf{the limit exists}$, but $\bf\lim\limits_{x \to 0}f(x)\ne f(0)$
we have a $\textbf{removable or point discontinuity at x=0}$
$\textbf{See graph}$
