let f(x)=\\(\\begin{cases}x - 1 & \\text{if }x<4 \\\\ 3 & \\text{if }4\\leq x\\leq8 \\\\ x + 1 & \\text{if…

let f(x)=\\(\\begin{cases}x - 1 & \\text{if }x<4 \\\\ 3 & \\text{if }4\\leq x\\leq8 \\\\ x + 1 & \\text{if }x>8\\end{cases}\\). a. find \\(\\lim_{x\\to4}f(x)\\). b. find \\(\\lim_{x\\to8}f(x)\\). select the correct choice and, if necessary, fill in the answer box to complete your choice. a. \\(\\lim_{x\\to4}=\\square\\) b. the limit is not -\\(\\infty\\) or \\(\\infty\\) and does not exist.

let f(x)=\\(\\begin{cases}x - 1 & \\text{if }x<4 \\\\ 3 & \\text{if }4\\leq x\\leq8 \\\\ x + 1 & \\text{if }x>8\\end{cases}\\). a. find \\(\\lim_{x\\to4}f(x)\\). b. find \\(\\lim_{x\\to8}f(x)\\). select the correct choice and, if necessary, fill in the answer box to complete your choice. a. \\(\\lim_{x\\to4}=\\square\\) b. the limit is not -\\(\\infty\\) or \\(\\infty\\) and does not exist.

Answer

Explanation:

Step1: Find left - hand limit as x→4

For (x < 4), (f(x)=x - 1). So, (\lim_{x\rightarrow4^{-}}f(x)=\lim_{x\rightarrow4^{-}}(x - 1)=4-1 = 3).

Step2: Find right - hand limit as x→4

For (4\leq x\leq8), (f(x)=3). So, (\lim_{x\rightarrow4^{+}}f(x)=\lim_{x\rightarrow4^{+}}3 = 3).

Step3: Determine (\lim_{x\rightarrow4}f(x))

Since (\lim_{x\rightarrow4^{-}}f(x)=\lim_{x\rightarrow4^{+}}f(x)=3), then (\lim_{x\rightarrow4}f(x)=3).

Step4: Find left - hand limit as x→8

For (4\leq x\leq8), (f(x)=3). So, (\lim_{x\rightarrow8^{-}}f(x)=\lim_{x\rightarrow8^{-}}3 = 3).

Step5: Find right - hand limit as x→8

For (x>8), (f(x)=x + 1). So, (\lim_{x\rightarrow8^{+}}f(x)=\lim_{x\rightarrow8^{+}}(x + 1)=8 + 1=9).

Step6: Determine (\lim_{x\rightarrow8}f(x))

Since (\lim_{x\rightarrow8^{-}}f(x)=3) and (\lim_{x\rightarrow8^{+}}f(x)=9), (\lim_{x\rightarrow8^{-}}f(x)\neq\lim_{x\rightarrow8^{+}}f(x)), so (\lim_{x\rightarrow8}f(x)) does not exist.

Answer:

a. A. (\lim_{x\rightarrow4}f(x)=3) b. B. The limit is not (-\infty) or (\infty) and does not exist.