## Calculus: Early Transcendentals 8th Edition

(a) (i) $\lim\limits_{x \to -2^+}\lfloor{x}\rfloor = -2$ (ii) $\lim\limits_{x \to -2}\lfloor{x}\rfloor$ does not exist. (iii)$\lim\limits_{x \to -2.4}\lfloor{x}\rfloor = -3$ (b) (i) $\lim\limits_{x \to n^-}\lfloor{x}\rfloor = n-1$ (ii) $\lim\limits_{x \to n^+}\lfloor{x}\rfloor = n$ (c) $\lim\limits_{x \to a}\lfloor{x}\rfloor$ exists if $a$ is not an integer. If $a$ is an integer, then the limit does not exist.
(a) (i) $\lim\limits_{x \to -2^+}\lfloor{x}\rfloor = \lim\limits_{x \to -2^+} -2 = -2$ (ii) $\lim\limits_{x \to -2^-}\lfloor{x}\rfloor = \lim\limits_{x \to -2^-} -3 = -3$ Since $\lim\limits_{x \to -2^-}\lfloor{x}\rfloor \neq \lim\limits_{x \to -2^+}\lfloor{x}\rfloor$, then $\lim\limits_{x \to -2}\lfloor{x}\rfloor$ does not exist. (iii)$\lim\limits_{x \to -2.4}\lfloor{x}\rfloor = \lim\limits_{x \to -2.4} -3 = -3$ (b) Suppose that $n$ is an integer. (i) $\lim\limits_{x \to n^-}\lfloor{x}\rfloor = \lim\limits_{x \to n^-} (n-1) = n-1$ (ii) $\lim\limits_{x \to n^+}\lfloor{x}\rfloor = \lim\limits_{x \to n^+} n = n$ (c) If $a$ is an integer, then $\lim\limits_{x \to a^-}\lfloor{x}\rfloor \neq \lim\limits_{x \to a^+}\lfloor{x}\rfloor$. Therefore, $\lim\limits_{x \to a}\lfloor{x}\rfloor$ does not exist if $a$ is an integer. If $a$ is not an integer, then $\lim\limits_{x \to a^-}\lfloor{x}\rfloor = \lim\limits_{x \to a^+}\lfloor{x}\rfloor = \lfloor{a}\rfloor$ Therefore, $\lim\limits_{x \to a}\lfloor{x}\rfloor$ exists if $a$ is not an integer.