hw5 the limit laws (target l4; §2.3)\nscore: 11/13 answered: 12/13\nquestion 13\nlet $f(x)=\begin{cases}-3x…

Explanation:

Step1: Find left - hand limit

We use the part of the piece - wise function for $x\leq1$. $\lim_{x\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{-}}(-3x + 5)$ Substitute $x = 1$ into $-3x + 5$: $-3(1)+5=2$

Step2: Find right - hand limit

We use the part of the piece - wise function for $x>1$. $\lim_{x\rightarrow1^{+}}f(x)=\lim_{x\rightarrow1^{+}}(-6x + 6)$ Substitute $x = 1$ into $-6x + 6$: $-6(1)+6=0$

Step3: Check if limit exists

Since $\lim_{x\rightarrow1^{-}}f(x)=2$ and $\lim_{x\rightarrow1^{+}}f(x)=0$, and $\lim_{x\rightarrow1^{-}}f(x)\neq\lim_{x\rightarrow1^{+}}f(x)$. The limit $\lim_{x\rightarrow1}f(x)$ does not exist.

Answer:

DNE