Answer
(a) $\lim_{x\to2}f(x)=0$
(b) $\lim_{x\to-3^+}f(x)=-2$
(c) $\lim_{x\to-3^-}f(x)=2$
(d) $\lim_{x\to-3}f(x)$ does not exist
(e) $\lim_{x\to0^+}f(x)=-1$
(f) $\lim_{x\to0^-}f(x)=\infty$
(g) $\lim_{x\to0}f(x)$ does not exist
(h) $\lim_{x\to\infty}f(x)=1$
(i) $\lim_{x\to-\infty}f(x)=0$
Work Step by Step
(a) $\lim_{x\to2}f(x)=0$
As $x$ approaches $2$, $f(x)$ approaches $0$ from both the left and the right side.
(b) $\lim_{x\to-3^+}f(x)=-2$
As $x$ approaches $-3$ from the right, $f(x)$ approaches $-2$.
(c) $\lim_{x\to-3^-}f(x)=2$
As $x$ approaches $-3$ from the left, $f(x)$ approaches $2$.
(d) $\lim_{x\to-3}f(x)$ does not exist
Since $\lim_{x\to-3^-}f(x)\ne\lim_{x\to-3^+}f(x)$, there is no single value that $f(x)$ would approach as $x\to-3$.
(e) $\lim_{x\to0^+}f(x)=-1$
As $x$ approaches $0$ from the right, $f(x)$ approaches $-1$.
(f) $\lim_{x\to0^-}f(x)=\infty$
As $x$ approaches $0$ from the left, $f(x)$ gets infinitely large, so we can say $f(x)$ approaches $\infty$.
(g) $\lim_{x\to0}f(x)$ does not exist
Since $\lim_{x\to0^-}f(x)\ne\lim_{x\to0^+}f(x)$, there is no single value that $f(x)$ would approach as $x\to0$.
(h) $\lim_{x\to\infty}f(x)=1$
As $x$ gets infinitely large, or approaches $\infty$, $f(x)$ approaches $1$.
(i) $\lim_{x\to-\infty}f(x)=0$
As $x$ gets infinitely small, or approaches $-\infty$, $f(x)$ approaches $0$.
