Answer
$\displaystyle\lim_{x\rightarrow 1^{-}} f(x)=3$
$\displaystyle\lim_{x\rightarrow 1^{+}} f(x)=3$
$\displaystyle\lim_{x\rightarrow 3^{-}} f(x)=-\infty$
$\displaystyle\lim_{x\rightarrow 3^{+}} f(x)=4$
$\displaystyle\lim_{x\rightarrow 5^{-}} f(x)=2$
$\displaystyle\lim_{x\rightarrow 5^{+}} f(x)=-3$
$\displaystyle\lim_{x\rightarrow 6^{-}} f(x)=\infty$
$\displaystyle\lim_{x\rightarrow 6^{+}} f(x)=\infty$
Work Step by Step
We have to determine the limits:
$\displaystyle\lim_{x\rightarrow 1\pm} f(x)$
$\displaystyle\lim_{x\rightarrow 3\pm} f(x)$
$\displaystyle\lim_{x\rightarrow 5\pm} f(x)$
$\displaystyle\lim_{x\rightarrow 6\pm} f(x)$
As we approach 1 from the left and the right, the value of $f(x)$ approaches 3:
$\displaystyle\lim_{x\rightarrow 1^{-}} f(x)=3$
$\displaystyle\lim_{x\rightarrow 1^{+}} f(x)=3$
As we approach 3 from the left, the value of $f(x)$ approaches $-\infty$, while as we approach 3 from the right, the value of $f(x)$ approaches $4$:
$\displaystyle\lim_{x\rightarrow 3^{-}} f(x)=-\infty$
$\displaystyle\lim_{x\rightarrow 3^{+}} f(x)=4$
As we approach 5 from the left, the value of $f(x)$ approaches $2$, while as we approach 5 from the right, the value of $f(x)$ approaches $-3$:
$\displaystyle\lim_{x\rightarrow 5^{-}} f(x)=2$
$\displaystyle\lim_{x\rightarrow 5^{+}} f(x)=-3$
As we approach 6 from the left, the value of $f(x)$ approaches $\infty$, while as we approach 6 from the right, the value of $f(x)$ approaches $\infty$:
$\displaystyle\lim_{x\rightarrow 6^{-}} f(x)=\infty$
$\displaystyle\lim_{x\rightarrow 6^{+}} f(x)=\infty$
