h(x)={[5x," for "x = -2]:} Find lim_(x rarr-2)h(x). Choose 1 answer: (A) -10 (B) -2 (C) 10

Define h(x): We need to find the limit of h(x) as x approaches -2. The function h(x) is defined piecewise: h(x) = { 5x for x = -2 } To find the limit as x approaches -2, we need to consider the value from both sides of -2.Limit from left side: First, let's find the limit from the left side (x approaching -2 from values less than -2). For x -2^-) h(x) = lim_(x -> -2^-) 5x = 5 * (-2) = -10.Limit from right side: Now, let's find the limit from the right side (x approaching -2 from values greater than or equal to -2). For x >= -2, h(x) = x^3 - 2. lim_(x -> -2^+) h(x) = lim_(x -> -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. lim_(x -> -2) h(x) = -10.

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$h(x)={[5x," for "x < -2],[x^(3)-2," for "x >= -2]:} Find lim_(x rarr-2)h(x). Choose 1 answer: (A) -10 (B) -2 (C) 10$

$h(x)=\begin{cases} 5x, & \text{for } x < -2 \ x^{3}-2, & \text{for } x \geq -2 \end{cases}$ $\newline$ Find $\lim_{x \to -2}h(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-10$ $\newline$ (B) $-2$ $\newline$ (C) $10$

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Algebra 1

Compare linear and exponential growth

Full solution

h(x)=\begin{cases} 5x, & \text{for } x < -2 \ x^{3}-2, & \text{for } x \geq -2 \end{cases}

\newline

Find

\lim_{x \to -2}h(x)

\newline

Choose

1

answer:

\newline

(A)

-10

\newline

(B)

-2

\newline

(C)

10

Define $h(x)$ : We need to find the limit of $h(x)$ as $x$ approaches $-2$ . The function $h(x)$ is defined piecewise: $\newline$ $h(x) = \begin{cases} 5x & \text{for } x < -2,\ x^3 - 2 & \text{for } x \geq -2 \end{cases}$ $\newline$ To find the limit as $x$ approaches $-2$ , we need to consider the value from both sides of $-2$ .
Limit from left side: First, let's find the limit from the left side $x$ approaching $-2$ from values less than $-2$ . For x < -2, $h(x) = 5x$ . $\lim_{x \to -2^-} h(x) = \lim_{x \to -2^-} 5x = 5 \times (-2) = -10$ .
Limit from right side: Now, let's find the limit from the right side $x$ approaching $-2$ from values greater than or equal to $-2$ . For $x \geq -2$ , $h(x) = x^3 - 2$ . $\lim_{x \to -2^+} h(x) = \lim_{x \to -2^+} (x^3 - 2) = (-2)^3 - 2 = -8 - 2 = -10$ .
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 $-10$ . $\newline$ $\lim_{x \to -2} h(x) = -10$ .