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

(a) $\lim_{x\to4}f(x)=2$ (b) $\lim_{x\to2^+}f(x)=-3$ (c) $\lim_{x\to2^-}f(x)=1$ (d) $\lim_{x\to2}f(x)$ does not exist (e) $\lim_{x\to-3^+}f(x)=\infty$ (f) $\lim_{x\to-3^-}f(x)=\infty$ (g) $\lim_{x\to-3}f(x)=\infty$ (h) $\lim_{x\to0^+}f(x)=\infty$ (i) $\lim_{x\to0^-}f(x)=-\infty$ (j) $\lim_{x\to0}f(x)$ does not exist (k) $\lim_{x\to\infty}f(x)=0$ (l) $\lim_{x\to-\infty}f(x)=-1$
(a) $\lim_{x\to4}f(x)=2$ As $x$ approaches $4$, $f(x)$ approaches $2$ from both the left and the right side. (b) $\lim_{x\to2^+}f(x)=-3$ As $x$ approaches $2$ from the right, $f(x)$ approaches $-3$. (c) $\lim_{x\to2^-}f(x)=1$ As $x$ approaches $2$ from the left, $f(x)$ approaches $1$. (d) $\lim_{x\to2}f(x)$ does not exist Since $\lim_{x\to2^-}f(x)\ne\lim_{x\to2^+}f(x)$, there is no single value that $f(x)$ would approach as $x\to2$. (e) $\lim_{x\to-3^+}f(x)=\infty$ As $x$ approaches $-3$ from the right, $f(x)$ gets infinitely large, or approaches $\infty$. (f) $\lim_{x\to-3^-}f(x)=\infty$ As $x$ approaches $-3$ from the left, $f(x)$ gets infinitely large, so we can say $f(x)$ approaches $\infty$. (g) $\lim_{x\to-3}f(x)=\infty$ As $f(x)$ approaches infinity as $x$ approaches $-3$ from both the right and the left sides, $f(x)$ approaches infinity as $x$ approaches $-3$. (h) $\lim_{x\to0^+}f(x)=\infty$ As $x$ approaches $0$ from the right, $f(x)$ gets infinitely large, or approaches $\infty$. (i) $\lim_{x\to0^-}f(x)=-\infty$ As $x$ approaches $0$ from the left, $f(x)$ gets infinitely small, so we can say $f(x)$ approaches $-\infty$. (j) $\lim_{x\to0}f(x)$ does not exist As $f(x)$ gets infinitely large as $x\to0^+$ while $f(x)$ gets infinitely small as $x\to0^-$, $f(x)$ does not approach any single value as $x\to0$, nor is there any symbol to denote this situation. So we say $\lim_{x\to0}f(x)$ does not exist. (k) $\lim_{x\to\infty}f(x)=0$ As $x$ gets infinitely large, or approaches $\infty$, $f(x)$ approaches $0$. (l) $\lim_{x\to-\infty}f(x)=-1$ As $x$ gets infinitely small, or approaches $-\infty$, $f(x)$ approaches $-1$.