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What is the value of $$\limsup_{x\to \pi} \sin(x)$$ ?

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Sine is continuous so this is the same as the limit, so again as sine is continuous, this is the same as the value of sine at $\pi$ which is $0$. – user45150 Feb 6 '13 at 17:48
If the solution provided by user45150 doesn't help you understand the problem, perhaps it would be best if you could describe what's confusing you. – Antonio Vargas Feb 6 '13 at 18:55
@user45150 Isn't it equal to 1? Why isn't $\pi/2$ in the neighborhood of $\pi$? – user1611823 Feb 6 '13 at 18:58
Which neighborhood of $\pi$? It has many, and $\pi/2$ is not in all of them. – Trevor Wilson Feb 6 '13 at 19:08

(Getting this question off the unanswered list...)

As user45150 wrote, the sine function is continuous, so $$\limsup_{x\to\pi} \sin(x) = \lim_{x\to\pi} \sin(x)= \sin(\pi) = 0.$$

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