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1 of the answers is not correct.
let $f(x) = \

$$\begin{cases}8 - x - x^2, & \\text{if } x \\leq 2 \\\\ 2x - 3, & \\text{if } x > 2 \\end{cases}$$

$
calculate the following limits. enter dne if the limit does not exist.
$\lim\limits_{x \to 2^-} f(x) = 6$ $\lim\limits_{x \to 2^+} f(x) = 1$ $\lim\limits_{x \to 2} f(x) = dne$
note: you can earn partial credit on this problem.
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Explanation:

Step1: Evaluate left-hand limit

Use $f(x)=8-x-x^2$ for $x\leq2$
$\lim_{x\to2^-}f(x)=8-2-(2)^2=8-2-4=2$

Step2: Confirm right-hand limit

Use $f(x)=2x-3$ for $x>2$
$\lim_{x\to2^+}f(x)=2(2)-3=4-3=1$

Step3: Check overall limit

Left $
eq$ right limit, so limit DNE.

Answer:

$\lim_{x\to2^-}f(x)=2$
$\lim_{x\to2^+}f(x)=1$
$\lim_{x\to2}f(x)=$ dne