let (f(x)=\begin{cases}2 - x - x^{2}&\text{if }xleq2\\2x - 5&\text{if }x>2end{cases}\ncalculate the…

let (f(x)=\begin{cases}2 - x - x^{2}&\text{if }xleq2\\2x - 5&\text{if }x>2end{cases}\ncalculate the following limits. enter \dne\ if the limit does not exist.\n(lim_{x\rightarrow2^{-}}f(x)=)\n(lim_{x\rightarrow2^{+}}f(x)=)\n(lim_{x\rightarrow2}f(x)=)\nquestion help: video message instructor

let (f(x)=\begin{cases}2 - x - x^{2}&\text{if }xleq2\\2x - 5&\text{if }x>2end{cases}\ncalculate the following limits. enter \dne\ if the limit does not exist.\n(lim_{x\rightarrow2^{-}}f(x)=)\n(lim_{x\rightarrow2^{+}}f(x)=)\n(lim_{x\rightarrow2}f(x)=)\nquestion help: video message instructor

Answer

Explanation:

Step1: Find left - hand limit

For $\lim_{x\rightarrow2^{-}}f(x)$, since $x\rightarrow2^{-}$ means $x < 2$, we use the function $f(x)=2 - x - x^{2}$. Substitute $x = 2$ into $2 - x - x^{2}$: [ \begin{align*} 2-2 - 2^{2}&=2-2 - 4\ &=- 4 \end{align*} ]

Step2: Find right - hand limit

For $\lim_{x\rightarrow2^{+}}f(x)$, since $x\rightarrow2^{+}$ means $x>2$, we use the function $f(x)=2x - 5$. Substitute $x = 2$ into $2x - 5$: [ \begin{align*} 2\times2-5&=4 - 5\ &=-1 \end{align*} ]

Step3: Determine the overall limit

Since $\lim_{x\rightarrow2^{-}}f(x)=-4$ and $\lim_{x\rightarrow2^{+}}f(x)=-1$, and $\lim_{x\rightarrow2^{-}}f(x)\neq\lim_{x\rightarrow2^{+}}f(x)$, then $\lim_{x\rightarrow2}f(x)$ does not exist (DNE).

Answer:

$\lim_{x\rightarrow2^{-}}f(x)=-4$ $\lim_{x\rightarrow2^{+}}f(x)=-1$ $\lim_{x\rightarrow2}f(x)=\text{DNE}$