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$f(x)={[-x^(2)+3x," if "x < 1],[2x+2," if "x >= 1]:} evaluate the limit lim_(x rarr1)f(x). If the limit does not exist, write DNE.$

$f(x)=\left\{\begin{array}{cl} -x^{2}+3 x & \text { if } x<1 \\ 2 x+2 & \text { if } x \geq 1 \end{array}\right.$ $\newline$ evaluate the limit $\lim _{x \rightarrow 1} f(x)$ . If the limit does not exist, write DNE.

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Algebra 2

Conjugate root theorems

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f(x)=\left\{\begin{array}{cl} -x^{2}+3 x & \text { if } x<1 \\ 2 x+2 & \text { if } x \geq 1 \end{array}\right.

\newline

evaluate the limit

\lim _{x \rightarrow 1} f(x)

. If the limit does not exist, write DNE.

Approach $1$ from Left: To evaluate the limit of a piecewise function as $x$ approaches a point, we need to consider the limit from both sides of that point. In this case, we need to find the limit of $f(x)$ as $x$ approaches $1$ from the left and from the right.
Approach $1$ from Right: First, let's find the limit as $x$ approaches $1$ from the left, which means we will use the part of the function that applies when $x < 1$ : $f(x) = -x^2 + 3x$ . $\newline$ $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (-x^2 + 3x)$ $\newline$ Now we substitute $x$ with $1$ . $\newline$ $\lim_{x \to 1^-} f(x) = -(1)^2 + 3(1) = -1 + 3 = 2$
Comparison and Conclusion: Next, we find the limit as $x$ approaches $1$ from the right, which means we will use the part of the function that applies when $x \geq 1$ : $f(x) = 2x + 2$ . $\newline$ $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x + 2)$ $\newline$ Now we substitute $x$ with $1$ . $\newline$ $\lim_{x \to 1^+} f(x) = 2(1) + 2 = 2 + 2 = 4$
Comparison and Conclusion: Next, we find the limit as $x$ approaches $1$ from the right, which means we will use the part of the function that applies when $x \geq 1$ : $f(x) = 2x + 2$ .
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x + 2)$
Now we substitute $x$ with $1$ .
$\lim_{x \to 1^+} f(x) = 2(1) + 2 = 2 + 2 = 4$ Now we compare the two one-sided limits. If they are equal, the limit exists and is equal to that value. If they are not equal, the limit does not exist (DNE).
From the left, we found the limit to be $2$ . From the right, we found the limit to be $4$ . Since these two values are not equal, the limit as $x$ approaches $1$ does not exist.