## Thomas' Calculus 13th Edition

$2$
In exercise 6, we found that $\displaystyle \lim_{x\rightarrow 1}f(x)=2.$ Also, $f(1)=1$. The function value at x=1 is different than the limit at x=1. Apply the Continuity Test (see p.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 $x\rightarrow c$). 3. $\displaystyle \lim_{x\rightarrow c}f(x)=f(c)$ (the limit equals the function value). Condition 3 is not satisfied, while the first two are. If we changed the value of f(1) to 2, then all three conditions of the Continuity Test would be satisfied, and f would be continuous at x=1.