Answer
(a) True (b) True (c) False (d) True (e) True (f) True
(g) False (h) False (i) False (j) False (k) True (l) False
Work Step by Step
(a) $\lim_{x\to-1^+}f(x)=1$
TRUE. We do see intuitively that as $x$ approaches $-1$ from the right, $f(x)$ gets arbitrarily close to $1$.
(b) $\lim_{x\to0^-}f(x)=0$
TRUE. We do see intuitively that as $x$ approaches $0$ from the left, $f(x)$ gets arbitrarily close to $0$.
(c) $\lim_{x\to0^-}f(x)=1$
FALSE. We showed in (b) that $\lim_{x\to0^-}f(x)=0$.
(d) $\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)$
TRUE. We do see intuitively that as $x$ approaches $0$ from the right, $f(x)$ gets arbitrarily close to $0$. Therefore, $\lim_{x\to0^+}f(x)=0=\lim_{x\to0^-}f(x)$
(e) $\lim_{x\to0}f(x)$ exists
TRUE. Since we proved in $(d)$ that $\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=0$, this means $\lim_{x\to0}f(x)$ exists and equals $0$ as well.
(f) $\lim_{x\to0}f(x)=0$
TRUE. $\lim_{x\to0}f(x)=\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=0$
(g) $\lim_{x\to0}f(x)=1$
FALSE. As shown in $(f)$, $\lim_{x\to0}f(x)=0$
(h) $\lim_{x\to1}f(x)=1$
FALSE. We can see intuitively:
- $\lim_{x\to1^-}f(x)=1$, because as $x$ approaches $1$ from the left, $f(x)$ gets arbitrarily close to $1$.
- $\lim_{x\to1^+}f(x)=0$, because as $x$ approaches $1$ from the right, $f(x)$ gets arbitrarily close to $0$.
Therefore, $\lim_{x\to1^+}f(x)\ne\lim_{x\to1^-}f(x)$, so $\lim_{x\to1}f(x)$ does not exist.
(i) $\lim_{x\to1}f(x)=0$
FALSE. As proved in $(h)$, $\lim_{x\to1}f(x)$ does not exist.
(j) $\lim_{x\to2^-}f(x)=2$
FALSE. We see intuitively that as $x$ approaches $2$ from the left, $f(x)$ gets arbitrarily close $0$, instead of $2$. So, $\lim_{x\to2^-}f(x)=0$
(k) $\lim_{x\to-1^-}f(x)$ does not exist.
TRUE. There are no values of $f(x)$ as $x$ approaches $-1$ from the left. In other words, $\lim_{x\to-1^-}f(x)$ does not exist.
(l) $\lim_{x\to2^+}f(x)=0$
FALSE. The graph stops at $x=2$, so there are no values of $f(x)$ as $x$ approaches $2$ from the right. In other words, $\lim_{x\to2^+}f(x)$ does not exist.
