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

Conditions for Continuity: To determine if h(x) is continuous at x=0, we need to check if the following three conditions are met: 1. h(0) is defined. 2. The limit of h(x) as x approaches 0 from the left (lim x->0^- h(x)) exists. 3. The limit of h(x) as x approaches 0 from the right (lim x->0^+ h(x)) exists and is equal to the limit from the left and to h(0).Finding h(0): First, we find h(0) using the definition of h(x) for x >= 0, which is h(x) = sqrt(x+pi). Substitute x = 0 into the function to get h(0) = sqrt(0+pi) = sqrt(pi).Limit from the Left: Next, we find the limit of h(x) as x approaches 0 from the left, which is the limit of sin(x) as x approaches 0. lim x->0^- sin(x) = sin(0) = 0.Limit from the Right: Now, we find the limit of h(x) as x approaches 0 from the right, which is the limit of sqrt(x+pi) as x approaches 0. lim x->0^+ sqrt(x+pi) = sqrt(0+pi) = sqrt(pi).Comparison of Limits: We compare the limits from the left and the right and the value of h(0). Since lim x->0^- sin(x) = 0 and lim x->0^+ sqrt(x+pi) = sqrt(pi), and h(0) = sqrt(pi), the limits are not equal. Therefore, h(x) is not continuous at x=0 because the limit from the left does not equal the limit from the right, nor does it equal h(0).

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Let

h(x)={[sin(x)," for "x < 0],[sqrt(x+pi)," for "x >= 0]:}
Is
h continuous at
x=0 ?
Choose 1 answer:
(A) Yes
(B)
No

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

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Intermediate Value Theorem

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Q. Let

\newline

h(x)=\left\{\begin{array}{ll} \sin (x) & \text { for } x<0 \\ \sqrt{x+\pi} & \text { for } x \geq 0 \end{array}\right.

\newline

h

continuous at

x=0

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B)

\mathrm{No}

Conditions for Continuity: To determine if $h(x)$ is continuous at $x=0$ , we need to check if the following three conditions are met: $\newline$ $1$ . $h(0)$ is defined. $\newline$ $2$ . The limit of $h(x)$ as $x$ approaches $0$ from the left ( $\lim_{x\to0^-} h(x)$ ) exists. $\newline$ $3$ . The limit of $h(x)$ as $x$ approaches $0$ from the right ( $x=0$ $0$ ) exists and is equal to the limit from the left and to $h(0)$ .
Finding $h(0)$ : First, we find $h(0)$ using the definition of $h(x)$ for $x \geq 0$ , which is $h(x) = \sqrt{x+\pi}$ . $\newline$ Substitute $x = 0$ into the function to get $h(0) = \sqrt{0+\pi} = \sqrt{\pi}$ .
Limit from the Left: Next, we find the limit of $h(x)$ as $x$ approaches $0$ from the left, which is the limit of $ext{sin}(x)$ as $x$ approaches $0$ . $\newline$ $ext{lim}_{x \to 0^-} ext{sin}(x) = ext{sin}(0) = 0.$
Limit from the Right: Now, we find the limit of $h(x)$ as $x$ approaches $0$ from the right, which is the limit of $\sqrt{x+\pi}$ as $x$ approaches $0$ .
$\lim_{x\to 0^+} \sqrt{x+\pi} = \sqrt{0+\pi} = \sqrt{\pi}$ .
Comparison of Limits: We compare the limits from the left and the right and the value of $h(0)$ . Since $\lim_{x \to 0^-} \sin(x) = 0$ and $\lim_{x \to 0^+} \sqrt{x+\pi} = \sqrt{\pi}$ , and $h(0) = \sqrt{\pi}$ , the limits are not equal. Therefore, $h(x)$ is not continuous at $x=0$ because the limit from the left does not equal the limit from the right, nor does it equal $h(0)$ .