consider the function g(x) = {-9, x < 11; 7, x >= 11}. what is the limit, if it exists? lim(x→11) g(x)

Explanation:

Step1: Find left - hand limit

We find $\lim_{x\rightarrow11^{-}}g(x)$. Since $x\rightarrow11^{-}$ means $x < 11$, $g(x)=-9$. So $\lim_{x\rightarrow11^{-}}g(x)=-9$.

Step2: Find right - hand limit

We find $\lim_{x\rightarrow11^{+}}g(x)$. Since $x\rightarrow11^{+}$ means $x\geq11$, $g(x) = 7$. So $\lim_{x\rightarrow11^{+}}g(x)=7$.

Step3: Check if limit exists

For $\lim_{x\rightarrow a}g(x)$ to exist, $\lim_{x\rightarrow a^{-}}g(x)=\lim_{x\rightarrow a^{+}}g(x)$. Here, $\lim_{x\rightarrow11^{-}}g(x)=-9$ and $\lim_{x\rightarrow11^{+}}g(x)=7$. Since $-9\neq7$, the limit does not exist.

Answer:

C. DNE