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$h(x)={[5x," for "x < -2],[x^(3)-2," for "x >= -2]:} Find lim_(x rarr-2)h(x). Choose 1 answer: (A) -10 (B) -2 (C) 10 (D) The limit doesn't exist.$

$h(x)=\left\{\begin{array}{ll} 5 x & \text { for } x<-2 \\ x^{3}-2 & \text { for } x \geq-2 \end{array}\right.$ $\newline$ Find $\lim _{x \rightarrow-2} h(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-10$ $\newline$ (B) $-2$ $\newline$ (C) $10$ $\newline$ (D) The limit doesn't exist.

Full solution

h(x)=\left\{\begin{array}{ll} 5 x & \text { for } x<-2 \\ x^{3}-2 & \text { for } x \geq-2 \end{array}\right.

\newline

Find

\lim _{x \rightarrow-2} h(x)

\newline

Choose

1

answer:

\newline

(A)

-10

\newline

(B)

-2

\newline

(C)

10

\newline

(D) The limit doesn't exist.

Consider Limits: To find the limit of the piecewise function $h(x)$ as $x$ approaches $-2$ , we need to consider the limit from both sides of $-2$ , that is, from the left and from the right.
Left Side Limit: First, let's find the limit from the left side $x$ approaching $-2$ from values less than $-2$ . For $x < -2$ , the function is defined as $h(x) = 5x$ . $\lim_{x \to -2^-} h(x) = \lim_{x \to -2^-} 5x = 5 \times (-2) = -10$ .
Right Side Limit: Now, let's find the limit from the right side $x$ approaching $-2$ from values greater than or equal to $-2$ . For $x \geq -2$ , the function is defined as $h(x) = x^3 - 2$ . $\lim_{x \to -2^+} h(x) = \lim_{x \to -2^+} (x^3 - 2) = (-2)^3 - 2 = -8 - 2 = -10$ .
Final Limit: Since the limit from the left side and the limit from the right side are equal, the limit of $h(x)$ as $x$ approaches $-2$ exists and is equal to $-10$ .