## Calculus with Applications (10th Edition)

$$f(x)=\frac{-5+x}{3x(3x+1)}$$ $x$-values where the given function is discontinuous. are $$x=0 , \quad x=-\frac{1}{3}$$ and $$f(0), \quad f(-\frac{1}{3})$$ do not exist, also $$\lim _{x \rightarrow { -\frac{1}{3}}} f(x) ,\quad \lim _{x \rightarrow 0} f(x)$$ are not exist
$$f(x)=\frac{-5+x}{3x(3x+1)}$$ This rational function is discontinuous wherever the denominator is zero. There is a discontinuity when $x=0$ and $x=-\frac{1}{3}$ . so the given function is discontinuous at $x=0$ and $x=-\frac{1}{3}$, therefore $f(0), f(-\frac{1}{3})$ do not exist. We investigate $\lim _{x \rightarrow 0} f(x)$ at $x=0$ first. From the left, where $x$-values are less than $0$, $$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(\frac{-5+x}{3x(3x+1)} \right)=\infty$$ From the right, where $x$-values are greater than 0, $$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(\frac{-5+x}{3x(3x+1)} \right)=-\infty$$ so, $\lim _{x \rightarrow 0} f(x)$ does not exist since $$\lim _{x \rightarrow 0^{+}} f(x) \ne \lim _{x \rightarrow 0^{-}}f(x)$$ Now, we investigate $\lim _{x \rightarrow -\frac{1}{3}} f(x)$ at $x=-\frac{1}{3}$ first. From the left, where $x$-values are less than $-\frac{1}{3}$, $$\lim _{x \rightarrow{ -\frac{1}{3}}^{-}} f(x)=\lim _{x \rightarrow { -\frac{1}{3}}^{-}}\left(\frac{-5+x}{3x(3x+1)} \right)=-\infty$$ From the right, where $x$-values are greater than $-\frac{1}{3}$, $$\lim _{x \rightarrow { -\frac{1}{3}}^{+}} f(x)=\lim _{x \rightarrow { -\frac{1}{3}}^{+}}\left(\frac{-5+x}{3x(3x+1)} \right)=\infty$$ so, $\lim _{x \rightarrow { -\frac{1}{3}}} f(x)$ does not exist since $$\lim _{x \rightarrow { -\frac{1}{3}}^{+}} f(x) \ne \lim _{x \rightarrow { -\frac{1}{3}}^{-}}f(x)$$ Hence, $x$-values where the function is discontinuous. are $x=0$ and $x=-\frac{1}{3}$ . $f(0), f(-\frac{1}{3})$ do not exist, also $\lim _{x \rightarrow { -\frac{1}{3}}} f(x)$, $\lim _{x \rightarrow 0} f(x)$ do not exist.