## Thomas' Calculus 13th Edition

$a.\quad$yes $b.\quad$yes $c.\quad$no $d.\quad$no
The graph of f(x) contains points $(x,f(x))$, where x belongs to the domain of f. $a.$ The point $(1,1)$ belongs to the graph of f, so $f(1)=1$ ( $f(1)$ exists) $b.$ As x approaches the value 1 from the left, f(x) approaches 2, and, as x approaches the value 1 from the right, f(x) approaches 2. The one-sided limits exist, and are equal, $\displaystyle \lim_{x\rightarrow 1^{-}}f(x)=\lim_{x\rightarrow 1^{+}}f(x)=2$, so, by Th.6 (sec 2.4), a limit at x=1 exists (and equals 2), $\displaystyle \lim_{x\rightarrow 1}f(x)=2$ $c.$ $\displaystyle \lim_{x\rightarrow 1}f(x)=2\ \ \neq\ \ f(1)=1$ (they are not equal) $d.$ Apply "Continuity test", pg.95 1. $f(c)$ exists ( $c$ lies in the domain of $f$). 2. $\displaystyle \lim_{x\rightarrow c}f(x)$ exists ( $f$ has a limit as $x\rightarrow c$). The first two conditions are satisfied at $x=1$ BUT the third one is not: 3. $\displaystyle \lim_{x\rightarrow c}f(x)=f(c)$ (the limit equals the function value). Therefore, f is not continuous at $x=1$.