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

$f(x)=\left\{\begin{array}{ll} \sqrt{7+x} & \text { for }-7 \leq a \\ x^{2}-5 & \text { for } x>-3 \end{array}\right.$ $\newline$ 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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Algebra 1

Domain and range of square root functions: equations

Full solution

f(x)=\left\{\begin{array}{ll} \sqrt{7+x} & \text { for }-7 \leq a \\ 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.

Given function and limit: We are given a piecewise function $f(x)$ and asked to find the limit as $x$ approaches $-3$ . The function is defined differently for $x$ values less than or equal to $-7$ and for $x$ values greater than $-3$ . To find the limit as $x$ approaches $-3$ , we need to consider the definition of the function for values of $x$ near $-3$ .
Consider the function for x > -3: Since we are approaching $-3$ , we need to look at the part of the function that is defined for $x$ values greater than $-3$ , which is $f(x) = x^2 - 5$ .
Substitute $-3$ into the function: To find the limit as $x$ approaches $-3$ from the right, we substitute $-3$ into the function $f(x) = x^2 - 5$ . $\lim_{x \to -3} f(x) = (-3)^2 - 5 = 9 - 5 = 4.$
Check limit from the left: We also need to check the limit as $x$ approaches $-3$ from the left. However, since the function is not defined for $x$ values less than $-7$ , we only need to consider the limit from the right.
Overall limit as $x$ approaches $-3$ : Since the limit from the right exists and there is no function defined to the left of $-3$ (up to $-7$ ), the overall limit as $x$ approaches $-3$ is the same as the limit from the right.
Final result: Therefore, the limit of $f(x)$ as $x$ approaches $-3$ is $4$ .