## 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) $\lim_{x\to2}f(x)$ does not exist FALSE. We can see intuitively: - $\lim_{x\to2^-}f(x)=1$, because as $x$ approaches $2$ from the left, $f(x)$ gets arbitrarily close to $1$. - $\lim_{x\to2^+}f(x)=1$, because as $x$ approaches $2$ from the right, $f(x)$ gets arbitrarily close to $1$. Therefore, because $\lim_{x\to2^+}f(x)=\lim_{x\to2^-}f(x)$, so $\lim_{x\to2}f(x)$ does exist and equals $1$. (c) $\lim_{x\to2}f(x)=2$ FALSE. We showed in (b) that $\lim_{x\to2}f(x)=1$. (d) $\lim_{x\to1^-}f(x)=2$ TRUE. We do see intuitively that as $x$ approaches $1$ from the left, $f(x)$ gets arbitrarily close to $2$. (e) $\lim_{x\to1^+}f(x)=1$ TRUE. We do see intuitively that as $x$ approaches $1$ from the right, $f(x)$ gets arbitrarily close to $1$. (f) $\lim_{x\to1}f(x)$ does not exist. TRUE. Since $\lim_{x\to1^-}f(x)\ne\lim_{x\to1^+}f(x)$, $\lim_{x\to1}f(x)$ does not exist. (g) $\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)$ TRUE. We do see intuitively that as $x$ approaches $0$ from both the left and the right, $f(x)$ gets arbitrarily close to $0$. So $\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)=0$ (h) $\lim_{x\to c}f(x)$ exists at every $c$ in the open interval $(-1,1)$ TRUE. At all $c$ in the open interval $(-1,1)$, as $x$ approaches $c$ from both the left and the right, $f(x)$ gets arbitrarily close to one value of $f(x)$ only, meaning $\lim_{x\to c}f(x)$ exists. (i) $\lim_{x\to c}f(x)$ exists at every $c$ in the open interval $(1,3)$ TRUE. At all $c$ in the open interval $(1,3)$, as $x$ approaches $c$ from both the left and the right, $f(x)$ gets arbitrarily close to one value of $f(x)$ only, meaning $\lim_{x\to c}f(x)$ exists. (j) $\lim_{x\to-1^-}f(x)=0$ FALSE. 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. (k) $\lim_{x\to3^+}f(x)$ does not exist. TRUE. There are no values of $f(x)$ as $x$ approaches $3$ from the right. In other words, $\lim_{x\to3^+}f(x)$ does not exist.