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 :)
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 :)
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.
This function is defined at points (kĻ)/2 where k = 0,1,2,3, so just apply the formula