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$f(x)={[sqrt(7+x)," for "-7 <= x <= -3],[x^(2)-5," for "x > -3]:} Find lim_(x rarr-3)f(x). Choose 1 answer: (A) -3 (B) 2 (C) 4 (D) The limit doesn't exist.$

$f(x)=\left\{\begin{array}{ll}\sqrt{7+x} & \text { for }-7 \leq x \leq-3 \\ x^{2}-5 & \text { for } x>-3\end{array}\right.$ $\newline$ Find $\lim _{x \rightarrow-3} f(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-3$ $\newline$ (B) $2$ $\newline$ (C) $4$ $\newline$ (D) The limit doesn't exist.

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Full solution

f(x)=\left\{\begin{array}{ll}\sqrt{7+x} & \text { for }-7 \leq x \leq-3 \\ x^{2}-5 & \text { for } x>-3\end{array}\right.

\newline

Find

\lim _{x \rightarrow-3} f(x)

\newline

Choose

1

answer:

\newline

(A)

-3

\newline

(B)

2

\newline

(C)

4

\newline

(D) The limit doesn't exist.

Find the Limit: We need to find the limit of the function $f(x)$ as $x$ approaches $-3$ . The function is defined piecewise, so we will consider the limit from both sides of $-3$ .
Limit from the Left: First, let's consider the limit from the left side as $x$ approaches $-3$ . For $x$ values less than or equal to $-3$ , the function is defined as $f(x) = \sqrt{7+x}$ .
Calculate Left Limit: We calculate the limit from the left by substituting $x = -3$ into the function $f(x) = \sqrt{7+x}$ . $\newline$ $\lim_{x \to -3^-} f(x) = \sqrt{7 + (-3)} = \sqrt{4} = 2$ .
Limit from the Right: Now, let's consider the limit from the right side as $x$ approaches $-3$ . For $x$ values greater than $-3$ , the function is defined as $f(x) = x^2 - 5$ .
Calculate Right Limit: We calculate the limit from the right by substituting $x = -3$ into the function $f(x) = x^2 - 5$ . $\lim_{x \to -3^+} f(x) = (-3)^2 - 5 = 9 - 5 = 4$ .
Compare One-Sided Limits: Since the left-hand limit as $x$ approaches $-3$ is $2$ and the right-hand limit as $x$ approaches $-3$ is $4$ , the two one-sided limits are not equal. Therefore, the limit of $f(x)$ as $x$ approaches $-3$ does not exist.