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

$g(x)$ is continuous on $(-\infty,\infty)$
$g(x)=\frac{x^2-x-6}{x-3}$ for $x\ne3$ and $g(x)=5$ for $x=3$ - Domain: $g(x)$ is defined on $(-\infty,\infty)$ 1) Is $g(x)$ continuous at $x=3$? $g(x)$ is continuous at $x=3$ if and only if $\lim_{x\to3}g(x)=g(3)=5$ $$\lim_{x\to3}g(x)=\lim_{x\to3}\frac{x^2-x-6}{x-3}$$ (as $x\to3$, $x$ in fact does not equal $3$, so we use the version of $g(x)$ as $x\ne3$) $$\lim_{x\to3}g(x)=\lim_{x\to3}\frac{(x-3)(x+2)}{x-3}=\lim_{x\to3}(x+2)$$ $$\lim_{x\to3}g(x)=3+2=5$$ That means $\lim_{x\to3}g(x)=g(3)$. $g(x)$ is continuous at $x=3$. 2) Is $g(x)$ continuous at other points? As $x\ne3$: - $\lim_{x\to c}(x^2-x-6)=c^2-c-6$ - $\lim_{x\to c}(x-3)=c-3$ So both $y=x^2-x-6$ and $y=x-3$ are continuous on $(-\infty,3)\cup(3,\infty)$ Applying Theorem 8, the quotient of 2 functions continuous at $x=c$ is also continuous at $x=c$ Therefore, $g(x)=\frac{x^2-x-6}{x-3}$ is continuous on $(-\infty,3)\cup(3,\infty)$. In conclusion, $g(x)$ is continuous on $(-\infty,\infty)$