How to calculate the limit limx→1⌊sin−1(x)⌋The function takes the value 1 at x=1 but while approaching 1 from the left side, it takes the value 0 at all points.The function is not defined at x&gt;1 so the right limit does not exist.The book mentions 1 as the answer as f(1)=1 but i don't understand why.

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How to calculate the limit limx→1⌊sin−1(x)⌋The function takes the value 1 at x=1 but while approaching 1 from the left side, it takes the value 0 at all points.The function is not defined at x&amp;gt;1 so the right limit does not exist.The book mentions 1 as the answer as f(1)=1 but i don&#039;t understand why.

goleuedigdp · Accepted Answer

Note that arcsin⁡(1)=π2, not arcsin⁡(1)=1; this means in particular that for values x to the immediate left of 1, also arcsin⁡(x)&amp;gt;1. Thus the function floor(arcsin⁡(x)) is continuous at 1.However, the answer in the book is bad if it doesn&#039;t mention the fact that the function is continuous near x-value 1 and just gives the answer.

helsinka04 · Accepted Answer

You are correct, the limit is only defined as:  limx→1−⌊arcsin⁡(x)⌋  and since for  x→1−  we have arcsin⁡(x)→π2&amp;gt;1   we have that the limit is 1.

How to calculate the limit \lim_{x\to1}\lfloor\sin^{-1}(x)\rfloor The function takes the v

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