lim_(x rarr(pi)/(2))sin(x)=? Choose 1 answer: (A) -1 (B) 0 (C) 1 (D) The limit doesn't exist.

Substitution of x: To find the limit of sin(x) as x approaches π/2, we can directly substitute the value of x with π/2 in the function sin(x), because sin(x) is continuous at x = π/2.Calculation of sin(π/2): Substitute x with π/2 in the function sin(x): lim_(x -> π/2) sin(x) = sin(π/2)Conclusion: Calculate the value of sin(π/2): sin(π/2) = 1Since we have calculated the value of the limit without any restrictions or conditions that would cause the limit to not exist, we can conclude that the limit exists and is equal to 1.

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Calculus

Find derivatives of logarithmic functions

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\lim _{x \rightarrow \frac{\pi}{2}} \sin (x)=?

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Choose

1

answer:

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(A)

-1

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(B)

0

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(C)

1

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(D) The limit doesn't exist.

Substitution of $x$ : To find the limit of $\sin(x)$ as $x$ approaches $\frac{\pi}{2}$ , we can directly substitute the value of $x$ with $\frac{\pi}{2}$ in the function $\sin(x)$ , because $\sin(x)$ is continuous at $x = \frac{\pi}{2}$ .
Calculation of $\sin(\frac{\pi}{2})$ : Substitute $x$ with $\frac{\pi}{2}$ in the function $\sin(x)$ : $\lim_{x \to \frac{\pi}{2}} \sin(x) = \sin(\frac{\pi}{2})$
Conclusion: Calculate the value of $\sin(\frac{\pi}{2})$ : $\sin(\frac{\pi}{2}) = 1$
Conclusion: Calculate the value of $\sin(\frac{\pi}{2})$ : $\sin(\frac{\pi}{2}) = 1$ Since we have calculated the value of the limit without any restrictions or conditions that would cause the limit to not exist, we can conclude that the limit exists and is equal to $1$ .