Hi i need help with problem 36.

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- Feb 8th 2013, 09:22 AMPetrusContinuous on a range
Hi i need help with problem 36.

- Feb 8th 2013, 09:31 AMjakncokeRe: Continuous on a range
sin(x) is a continous function, cos(x) is a continous function, so you only need to verify that at $\displaystyle x = \frac{\pi}{4} $, $\displaystyle sin(\frac{\pi}{4}) = cos(\frac{\pi}{4}) $. Why is this? Basically even though $\displaystyle \frac{\pi}{4} $ is not in the domain of sin(x), $\displaystyle lim_{x^{-} \to \frac{\pi}{4}} f(x) \to cos(\frac{\pi}{4}) $ $\displaystyle lim_{x^{-} \to \frac{\pi}{4}} f(x) = lim_{x^{-} \to \frac{\pi}{4}} sin(x)$ Since sin(x) is continous $\displaystyle lim_{x^{-} \to \frac{\pi}{4}} sin(x) = sin(\frac{\pi}{4}) $

In a gist, if you come at $\displaystyle x = \frac{\pi}{4} $ from the left, it should be the same as if you come at it from the right. The only point at which problem can arise is at $\displaystyle x = \frac{\pi}{4} $, which is why you need to check there.