Which of the following functions are continuous at x=0 ? g(x)=cot(x) h(x)=(1)/(x^(2)) Choose 1 answer: (A) g only (B) h only (C) Both g and h (D) Neither g nor h

Check Continuity at x=0: To determine if the functions are continuous at x=0, we need to check if they are defined at x=0 and if their limits as x approaches 0 exist and are equal to the function value at x=0.Consider g(x) = cot(x): Let's first consider g(x) = cot(x). The cotangent function is the reciprocal of the tangent function, which means cot(x) = 1/tan(x). Since the tangent of 0 is 0, cot(x) has a vertical asymptote at x=0, which means it is not defined at x=0. Therefore, g(x) is not continuous at x=0.Consider h(x) = 1/x^2: Now let's consider h(x) = 1/x^2. As x approaches 0, the denominator x^2 approaches 0, which makes the function value approach infinity. Therefore, the limit of h(x) as x approaches 0 does not exist, and h(x) is not continuous at x=0.

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Which of the following functions are continuous at
x=0 ?

g(x)=cot(x)

h(x)=(1)/(x^(2))
Choose 1 answer:
(A)
g only
(B)
h only
(C) Both
g and
h
(D) Neither
g nor
h

Which of the following functions are continuous at $x=0$ ? $\newline$ $g(x)=\cot (x)$ $\newline$ $h(x)=\frac{1}{x^{2}}$ $\newline$ Choose $1$ answer: $\newline$ (A) $g$ only $\newline$ (B) $h$ only $\newline$ (C) Both $g$ and $h$ $\newline$ (D) Neither $g$ nor $h$

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Q. Which of the following functions are continuous at

x=0

\newline

g(x)=\cot (x)

\newline

h(x)=\frac{1}{x^{2}}

\newline

Choose

1

answer:

\newline

(A)

g

only

\newline

(B)

h

only

\newline

g

and

h

\newline

(D) Neither

g

nor

h

Check Continuity at $x=0$ : To determine if the functions are continuous at $x=0$ , we need to check if they are defined at $x=0$ and if their limits as $x$ approaches $0$ exist and are equal to the function value at $x=0$ .
Consider $g(x) = \cot(x)$ : Let's first consider $g(x) = \cot(x)$ . The cotangent function is the reciprocal of the tangent function, which means $\cot(x) = \frac{1}{\tan(x)}$ . Since the tangent of $0$ is $0$ , $\cot(x)$ has a vertical asymptote at $x=0$ , which means it is not defined at $x=0$ . Therefore, $g(x)$ is not continuous at $x=0$ .
Consider $h(x) = \frac{1}{x^2}$ : Now let's consider $h(x) = \frac{1}{x^2}$ . As $x$ approaches $0$ , the denominator $x^2$ approaches $0$ , which makes the function value approach infinity. Therefore, the limit of $h(x)$ as $x$ approaches $0$ does not exist, and $h(x)$ is not continuous at $h(x) = \frac{1}{x^2}$ $0$ .