Let h(x)={[e^(x+2)," for "x -2]:} Is h continuous at x=-2 ? Choose 1 answer: (A) Yes (B) No

Check Definition: To determine if h(x) is continuous at x=-2, we need to check if the following three conditions are met: 1. h(x) is defined at x=-2. 2. The limit of h(x) as x approaches -2 from the left is equal to the limit of h(x) as x approaches -2 from the right. 3. The limit of h(x) as x approaches -2 is equal to the value of h(x) at x=-2.Find Left Limit: First, let's check if h(x) is defined at x=-2. Since h(x) has a piece for x <= -2, which is e^(x+2), we can substitute x=-2 into this piece to find h(-2). h(-2) = e^(-2+2) = e^0 = 1. So, h(x) is defined at x=-2.Find Right Limit: Next, we need to find the limit of h(x) as x approaches -2 from the left. Since the function for x -2 is (x+2)^e. As x approaches -2 from the right, the expression (x+2)^e approaches 0^e. Since any positive number to the power of e is positive, and 0^e is 0, the limit as x approaches -2 from the right is 0.Conclusion: Since the limit of h(x) as x approaches -2 from the left is 1, and the limit as x approaches -2 from the right is 0, the two limits are not equal. Therefore, h(x) is not continuous at x=-2 because the limits from the left and right do not match.

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$Let h(x)={[e^(x+2)," for "x <= -2],[(x+2)^(e)," for "x > -2]:} Is h continuous at x=-2 ? Choose 1 answer: (A) Yes (B) No$

Let $\newline$ $h(x)=\left\{\begin{array}{ll} e^{x+2} & \text { for } x \leq-2 \\ (x+2)^{e} & \text { for } x>-2 \end{array}\right.$ $\newline$ Is $h$ continuous at $x=-2$ ? $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) $\mathrm{No}$

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Q. Let

\newline

h(x)=\left\{\begin{array}{ll} e^{x+2} & \text { for } x \leq-2 \\ (x+2)^{e} & \text { for } x>-2 \end{array}\right.

\newline

h

continuous at

x=-2

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B)

\mathrm{No}

Check Definition: To determine if $h(x)$ is continuous at $x=-2$ , we need to check if the following three conditions are met: $\newline$ $1$ . $h(x)$ is defined at $x=-2$ . $\newline$ $2$ . The limit of $h(x)$ as $x$ approaches $-2$ from the left is equal to the limit of $h(x)$ as $x$ approaches $-2$ from the right. $\newline$ $3$ . The limit of $h(x)$ as $x$ approaches $-2$ is equal to the value of $h(x)$ at $x=-2$ .
Find Left Limit: First, let's check if $h(x)$ is defined at $x=-2$ . Since $h(x)$ has a piece for $x \leq -2$ , which is $e^{(x+2)}$ , we can substitute $x=-2$ into this piece to find $h(-2)$ . $\newline$ $h(-2) = e^{(-2+2)} = e^0 = 1$ . $\newline$ So, $h(x)$ is defined at $x=-2$ .
Find Right Limit: Next, we need to find the limit of $h(x)$ as $x$ approaches $-2$ from the left. Since the function for $x \leq -2$ is $e^{(x+2)}$ , the limit as $x$ approaches $-2$ from the left is also $e^0 = 1$ .
Compare Limits: Now, we need to find the limit of $h(x)$ as $x$ approaches $-2$ from the right. The function for x > -2 is $(x+2)^e$ . As $x$ approaches $-2$ from the right, the expression $(x+2)^e$ approaches $0^e$ . Since any positive number to the power of $e$ is positive, and $0^e$ is $x$ $1$ , the limit as $x$ approaches $-2$ from the right is $x$ $1$ .
Conclusion: Since the limit of $h(x)$ as $x$ approaches $-2$ from the left is $1$ , and the limit as $x$ approaches $-2$ from the right is $0$ , the two limits are not equal. Therefore, $h(x)$ is not continuous at $x=-2$ because the limits from the left and right do not match.