Question

let $f(x)=\begin{cases}10 - x - x^{2}&\text{if }xleq1\\2x - 1&\text{if }x>1end{cases}$ use a graph to determine the following limits. enter dne if the limit does not exist. $lim_{x
ightarrow1^{-}}f(x)=$ $lim_{x
ightarrow1^{+}}f(x)=$ $lim_{x
ightarrow1}f(x)=$ question help: message instructor

Explanation:

Step1: Find left - hand limit

For $x\to1^{-}$, we use $f(x)=10 - x - x^{2}$. Substitute $x = 1$ into $10 - x - x^{2}$: $10-1 - 1^{2}=8$.

Step2: Find right - hand limit

For $x\to1^{+}$, we use $f(x)=2x - 1$. Substitute $x = 1$ into $2x - 1$: $2\times1-1 = 1$.

Step3: Determine overall limit

Since $\lim_{x\to1^{-}}f(x)=8$ and $\lim_{x\to1^{+}}f(x)=1$, and $8
eq1$, $\lim_{x\to1}f(x)$ does not exist.

Answer:

$\lim_{x\to1^{-}}f(x)=8$
$\lim_{x\to1^{+}}f(x)=1$
$\lim_{x\to1}f(x)=DNE$