let\n$f(x)=\\begin{cases}x^{2}+1&\\text{if }x < 2\\\\(x - 3)^{2}&\\text{if }x\\geq2\\end{cases}$\n(a) find…

let\n$f(x)=\\begin{cases}x^{2}+1&\\text{if }x < 2\\\\(x - 3)^{2}&\\text{if }x\\geq2\\end{cases}$\n(a) find the following limits. (if an answer does not exist, enter dne.)\n$\\lim_{x\\to2^{-}}f(x)=$\n$\\lim_{x\\to2^{+}}f(x)=$\n(b) does $\\lim_{x\\to2}f(x)$ exist?\nyes\nno\n(c) sketch the graph of $f$.

let\n$f(x)=\\begin{cases}x^{2}+1&\\text{if }x < 2\\\\(x - 3)^{2}&\\text{if }x\\geq2\\end{cases}$\n(a) find the following limits. (if an answer does not exist, enter dne.)\n$\\lim_{x\\to2^{-}}f(x)=$\n$\\lim_{x\\to2^{+}}f(x)=$\n(b) does $\\lim_{x\\to2}f(x)$ exist?\nyes\nno\n(c) sketch the graph of $f$.

Answer

Explanation:

Step1: Find left - hand limit

For $\lim_{x\rightarrow2^{-}}f(x)$, since $x\rightarrow2^{-}$ means $x < 2$, we use $f(x)=x^{2}+1$. Substitute $x = 2$ into $x^{2}+1$. So, $\lim_{x\rightarrow2^{-}}f(x)=2^{2}+1=5$.

Step2: Find right - hand limit

For $\lim_{x\rightarrow2^{+}}f(x)$, since $x\rightarrow2^{+}$ means $x\geq2$, we use $f(x)=(x - 3)^{2}$. Substitute $x = 2$ into $(x - 3)^{2}$. So, $\lim_{x\rightarrow2^{+}}f(x)=(2 - 3)^{2}=1$.

Step3: Determine if the limit exists

The limit $\lim_{x\rightarrow2}f(x)$ exists if and only if $\lim_{x\rightarrow2^{-}}f(x)=\lim_{x\rightarrow2^{+}}f(x)$. Since $5\neq1$, $\lim_{x\rightarrow2}f(x)$ does not exist.

Answer:

(a) $\lim_{x\rightarrow2^{-}}f(x)=5$ $\lim_{x\rightarrow2^{+}}f(x)=1$ (b) No (c) To sketch the graph:

  • For $y=x^{2}+1$ when $x < 2$, it is a parabola opening upwards with vertex at $(0,1)$. The point $(2,5)$ is a hole (since $x<2$).
  • For $y=(x - 3)^{2}$ when $x\geq2$, it is a parabola opening upwards with vertex at $(3,0)$. The point $(2,1)$ is a solid - dot.