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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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    What limit are you trying to find?2011-11-18
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    $(k\pi)/2$ where k = 0,1,2,3 (four limits)2011-11-18
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    Note the Floor of $\sin(x)$ is -1 if $\sin x<1$ and is 0 if $\sin x>0$.2011-11-18
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    @David: Unless it's +1.2011-11-18
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    @David Indeed, thanks.2011-11-18
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    @DavidMitra, Thanks but I still don't see how to use that information2011-11-18

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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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    Hmm, Thanks but I fail to understand how you can do that since $\lfloor{sinx}\rfloor$ is not continues at $\pi/2$2011-11-18
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    That doesn't come into play; when taking limits, it does not matter what the function does at the limit point. $\lim_{x\rightarrow a} f(x)=L$ if $f(x)$ can be made as close to $L$ as desired by taking $x$ sufficiently close to, $\it\ but\ different\ from\ $ $a$.2011-11-18
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    Remember, in the definition of limits, the value of the function at the point $a$ doesn't matter at all, undefined or something weird, whatever. The limit is decided by the values in a small open set around $a$, not including $a$. So even though the value changes suddenly at $\pi/2$, the values around $\pi/2$ don't change so suddenly. Note, if a function is continuous at $a$, we have $ f(a) = \lim_{x\to a} f(x) . $ If it is not continuous, the limit may still exist, it just won't equal $f(a).$2011-11-18
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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 help2011-11-18
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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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    It is defined but It's not continues so I can't just apply it.. Or can I?2011-11-18
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    @Jason, You can't. A definition of continuity at $x=a$ is that $ \lim_{x\to a} f(x) = f(a).$ and since the floor function is not continuous at integers, we can't simply plug it in. Thinking about or drawing a graph of sin x, think about the limits from the left and the right separately. Do they agree?2011-11-18
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    @RagibZaman yep I agree that's exactly what I did in my head :). But that's why I am stuck :( and David's tip doesn't seem to help me2011-11-18
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    I had a student who liked to say "... and then you just *do the math* " !2011-11-18