Answer
(a) $f(-1)$ exists and equals $0$.
(b) $\lim_{x\to-1^+}f(x)$ exists and equals $0$.
(c) $\lim_{x\to-1^+}f(x)=f(-1)=0$
(d) $f$ is continuous at $x=-1$ (from the right)
Work Step by Step
(a) As $x=-1$ lies in the domain $[-1,0)$, we would employ the function $f(x)=x^2-1$.
So, $f(-1)=(-1)^2-1=1-1=0$
From the graph, we can also see that $f(-1)$ exists and equals $0$.
(b) As $x$ approaches $-1$ from the right, we would again employ the function $f(x)=x^2-1$ for the domain $[-1,0)$
So, $\lim_{x\to-1^+}f(x)=\lim_{x\to-1^+}(x^2-1)=(-1)^2-1=1-1=0$
$\lim_{x\to-1^+}f(x)$, thus, exists and equals $0$.
(c) From the results in (a) and (b), $\lim_{x\to-1^+}f(x)=f(-1)=0$
(d) Because $\lim_{x\to-1^+}f(x)=f(-1)=0$, according to definition, $f$ is continuous at $x=-1$ (from the right to be exact).
