## University Calculus: Early Transcendentals (3rd Edition)

(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.