consider the function $g(x)=\begin{cases}-9,x < 11\\7,xgeq11end{cases}$. what is $lim_{x\rightarrow11}g(x)$…

Explanation:

Step1: Find left - hand limit

We find $\lim_{x\rightarrow11^{-}}g(x)$. When $x\rightarrow11^{-}$, $x < 11$, so $g(x)=-9$. Then $\lim_{x\rightarrow11^{-}}g(x)=-9$.

Step2: Find right - hand limit

We find $\lim_{x\rightarrow11^{+}}g(x)$. When $x\rightarrow11^{+}$, $x\geq11$, so $g(x) = 7$. Then $\lim_{x\rightarrow11^{+}}g(x)=7$.

Step3: Check if limit exists

Since $\lim_{x\rightarrow11^{-}}g(x)=-9$ and $\lim_{x\rightarrow11^{+}}g(x)=7$, and $-9\neq7$, the two - sided limit $\lim_{x\rightarrow11}g(x)$ does not exist.

Answer:

DNE