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I am trying to solve a limit for the function $\cos(x/2)-\lfloor\sin x\rfloor$ but the floor function seems to confuse me.

What can I do to deal with this?

Thanks a-lot for the help :)

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    @DavidMitra, Thanks but I still don't see how to use that information – 2011-11-18

2 Answers 2

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First $ \lfloor\sin x\rfloor =\cases{ 1,& \sin x=1 \cr 0,& 1>\sin x>0\cr -1,& \sin x <0 } $

For $k=0$, you're computing $ \lim_{x\rightarrow 0} (\cos(x/2) - \lfloor\sin x\rfloor). $ Now $\lim\limits_{x\rightarrow 0} \cos(x/2)=1$.

But $\lim\limits_{x\rightarrow 0} \lfloor\sin x\rfloor $ does not exist, since $\lim\limits_{x\rightarrow0^+}\lfloor\sin x\rfloor=0$ and $\lim\limits_{x\rightarrow0^-}\lfloor\sin x\rfloor=-1$.

From the previous two observations, it follows that $\lim\limits_{x\rightarrow 0} (\cos(x/2) - \lfloor\sin x\rfloor)$ does not exist.

For $k=1$, you're computing $ \lim_{x\rightarrow \pi/2} (\cos(x/2) - \lfloor\sin x\rfloor). $ As $\lim\limits_{x\rightarrow \pi/2} \cos(x/2)=\sqrt2/2$ and $\lim\limits_{x\rightarrow \pi/2} \lfloor\sin x\rfloor=0, $ it follows that $ \lim\limits_{x\rightarrow 0} (\cos(x/2) - \lfloor\sin x\rfloor)=\sqrt2/2$.

I'll leave the other two limits for you.

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    :) Sorry I forgot that the functions do not have to be continues to apply arithmetic rules of limits - they just have to be defined :) Thanks for the help – 2011-11-18
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This function is defined at points (kπ)/2 where k = 0,1,2,3, so just apply the formula

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    I had a student who liked to say "... and then you just *do the math* " ! – 2011-11-18