Let h(x)={[e^(2x)," for "x = 0]:} Is h continuous at x=0 ? Choose 1 answer: (A) Yes (B) No

Check Left-hand Limit: To determine if h(x) is continuous at x=0, we need to check if the left-hand limit as x approaches 0 from the left (x = 0), and both of these limits must be equal to the value of the function at x=0.Calculate Left-hand Limit: First, let's find the left-hand limit of h(x) as x approaches 0 from the left. Since for x 0-) h(x) = lim (x -> 0-) e^(2x) = e^(2*0) = e^0 = 1.Calculate Right-hand Limit: Next, we need to find the right-hand limit of h(x) as x approaches 0 from the right. For x >= 0, h(x) = e^(5x), so we calculate the limit of e^(5x) as x approaches 0 from the right. lim (x -> 0+) h(x) = lim (x -> 0+) e^(5x) = e^(5*0) = e^0 = 1.Check Value at x=0: Now, we need to check the value of the function at x=0. Since x=0 is included in the interval for x >= 0, we use the definition of h(x) for x >= 0, which is h(x) = e^(5x). h(0) = e^(5*0) = e^0 = 1.Conclusion: Since the left-hand limit as x approaches 0, the right-hand limit as x approaches 0, and the value of the function at x=0 are all equal to 1, we can conclude that h(x) is continuous at x=0.

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$Let h(x)={[e^(2x)," for "x < 0],[e^(5x)," for "x >= 0]:} Is h continuous at x=0 ? Choose 1 answer: (A) Yes (B) No$

Let $\newline$ $h(x)=\left\{\begin{array}{ll} e^{2 x} & \text { for } x<0 \\ e^{5 x} & \text { for } x \geq 0 \end{array}\right.$ $\newline$ Is $h$ continuous at $x=0$ ? $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) No

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Q. Let

\newline

h(x)=\left\{\begin{array}{ll} e^{2 x} & \text { for } x<0 \\ e^{5 x} & \text { for } x \geq 0 \end{array}\right.

\newline

h

continuous at

x=0

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B) No

Check Left-hand Limit: To determine if $h(x)$ is continuous at $x=0$ , we need to check if the left-hand limit as $x$ approaches $0$ from the left (x < 0) is equal to the right-hand limit as $x$ approaches $0$ from the right ( $x \geq 0$ ), and both of these limits must be equal to the value of the function at $x=0$ .
Calculate Left-hand Limit: First, let's find the left-hand limit of $h(x)$ as $x$ approaches $0$ from the left. Since for x < 0, $h(x) = e^{2x}$ , we need to calculate the limit of $e^{2x}$ as $x$ approaches $0$ from the left. $\newline$ $\lim_{x \to 0^-} h(x) = \lim_{x \to 0^-} e^{2x} = e^{2\cdot0} = e^0 = 1.$
Calculate Right-hand Limit: Next, we need to find the right-hand limit of $h(x)$ as $x$ approaches $0$ from the right. For $x \geq 0$ , $h(x) = e^{5x}$ , so we calculate the limit of $e^{5x}$ as $x$ approaches $0$ from the right. $\newline$ $\lim_{x \to 0+} h(x) = \lim_{x \to 0+} e^{5x} = e^{5\cdot0} = e^0 = 1.$
Check Value at $x=0$ : Now, we need to check the value of the function at $x=0$ . Since $x=0$ is included in the interval for $x \geq 0$ , we use the definition of $h(x)$ for $x \geq 0$ , which is $h(x) = e^{5x}$ . $\newline$ $h(0) = e^{5\cdot 0} = e^0 = 1$ .
Conclusion: Since the left-hand limit as $x$ approaches $0$ , the right-hand limit as $x$ approaches $0$ , and the value of the function at $x=0$ are all equal to $1$ , we can conclude that $h(x)$ is continuous at $x=0$ .