Question

question 5. version 1*/1. score: 0/3 let $f(x)=\begin{cases}10 - x - x^{2}&\text{if }xleq2\\2x - 3&\text{if }x > 2end{cases}$ calculate the following limits. enter \dne\ if the limit does not exist. $lim_{x
ightarrow2^{-}}f(x)=$ $lim_{x
ightarrow2^{+}}f(x)=$ $lim_{x
ightarrow2}f(x)=$ score: n/a/1 n/a/1 n/a/1 had help available: video question 6. version 1*/1. score: 0/2 the graph below is the function $f(x)$

Explanation:

Step1: Calculate left - hand limit

For $\lim_{x
ightarrow2^{-}}f(x)$, since $x
ightarrow2^{-}$ (approaching 2 from the left), we use $f(x)=10 - x - x^{2}$. Substitute $x = 2$ into $10 - x - x^{2}$: $10-2 - 2^{2}=10 - 2-4 = 4$.

Step2: Calculate right - hand limit

For $\lim_{x
ightarrow2^{+}}f(x)$, since $x
ightarrow2^{+}$ (approaching 2 from the right), we use $f(x)=2x - 3$. Substitute $x = 2$ into $2x - 3$: $2\times2-3=4 - 3=1$.

Step3: Calculate overall limit

The overall limit $\lim_{x
ightarrow2}f(x)$ exists if and only if $\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{+}}f(x)$. Since $4
eq1$, $\lim_{x
ightarrow2}f(x)$ does not exist (DNE).

Answer:

$\lim_{x
ightarrow2}f(x)=$ DNE
$\lim_{x
ightarrow2^{-}}f(x)=4$
$\lim_{x
ightarrow2^{+}}f(x)=1$