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

Check left-hand limit: To determine if g(x) is continuous at x=0, we need to check if the left-hand limit as x approaches 0 from the negative side is equal to the right-hand limit as x approaches 0 from the positive side, and both of these limits must be equal to the value of g(0).Calculate left-hand limit: First, we calculate the left-hand limit, which is the limit of g(x) as x approaches 0 from the negative side. Since g(x) = sin(x) for x = 0, we need to find the limit of x^2 as x approaches 0 from the right. \[\lim_{x \to 0^+} x^2 = 0^2 = 0\]Calculate right-hand limit: Now, we need to check the value of g(x) at x=0. According to the definition of g(x), for x >= 0, g(x) = x^2. Therefore, g(0) = 0^2 = 0.Check value at x=0: Since the left-hand limit, right-hand limit, and the value of g(0) are all equal to 0, we can conclude that g(x) is continuous at x=0.

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

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

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Intermediate Value Theorem

Full solution

Q. Let

\newline

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

\newline

g

continuous at

x=0

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B) No

Check left-hand limit: To determine if $g(x)$ is continuous at $x=0$ , we need to check if the left-hand limit as $x$ approaches $0$ from the negative side is equal to the right-hand limit as $x$ approaches $0$ from the positive side, and both of these limits must be equal to the value of $g(0)$ .
Calculate left-hand limit: First, we calculate the left-hand limit, which is the limit of g(x) as x approaches $0$ from the negative side. Since g(x) = sin(x) for x < $0$ , we need to find the limit of sin(x) as x approaches $0$ from the left. $\newline$ $\lim_{x \to 0^-} \sin(x) = \sin(0) = 0$
Check right-hand limit: Next, we calculate the right-hand limit, which is the limit of g(x) as x approaches $0$ from the positive side. Since g(x) = x^ $2$ for x >= $0$ , we need to find the limit of x^ $2$ as x approaches $0$ from the right. $\newline$ $\lim_{x \to 0^+} x^2 = 0^2 = 0$
Calculate right-hand limit: Now, we need to check the value of $g(x)$ at $x=0$ . According to the definition of $g(x)$ , for $x \geq 0$ , $g(x) = x^2$ . Therefore, $g(0) = 0^2 = 0$ .
Check value at $x=0$ : Since the left-hand limit, right-hand limit, and the value of $g(0)$ are all equal to $0$ , we can conclude that $g(x)$ is continuous at $x=0$ .