# How to calculate the limit $\lim_{x\to1}\lfloor\sin^{-1}(x)\rfloor$?

The question is about finding $$\lim_{x\to1} f(x)$$ where $$f(x) = \lfloor\sin^{-1}(x)\rfloor$$

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 $x>1$ so the right limit does not exist.

The book mentions $1$ as the answer as $f(1)=1$ but i don't understand why.

What is the meaning of a limit?

Note that $\arcsin(1) = \frac\pi2$, not $\arcsin(1) = 1$; this means in particular that for values $x$ to the immediate left of $1$, also $\arcsin(x) > 1$. Thus the function $\mathrm{floor}(\arcsin(x))$ is continuous at $1$.

However, the answer in the book is bad if it doesn't mention the fact that the function is continuous near $x$-value 1 and just gives the answer.

• Sorry, I got confused and thought arcsin(1) = 1. Sorry and thanks. – Harshit Joshi Sep 3 '18 at 13:23
• But what if the question was about $lim_{x->1}floor(x)$ $f: [0,1]->{0,1}$ – Harshit Joshi Sep 3 '18 at 13:28
• Do you mean the case where $f(x) = \mathrm{floor}(x)$? In that case the correct answer would be $0$: for any $\epsilon > 0$, take $\delta = 1/2$, and then $0 < |x - 1| < \delta$ (which just means $1/2 < x < 1$) implies that $|f(x) - 0| = |0 - 0| = 0 < \epsilon$, as we needed to prove. – Mees de Vries Sep 3 '18 at 13:30
• Thanks for answering my stupid question – Harshit Joshi Sep 3 '18 at 13:32

You are correct, the limit is only defined as:

$$\lim_{x\to 1^-} \bigl\lfloor \arcsin (x)\bigr\rfloor$$

and since for $x\to 1^-$ we have $\arcsin (x)\to \frac{\pi}2 \gt1$, we have that the limit is $1$.

• The ordinary limit is also defined. – Yves Daoust Sep 3 '18 at 13:47
• @YvesDaoust Yes of course what I mean is that we are taking $x\to 1^-$ even if we are allowed to write $x\to 1$. – user Sep 3 '18 at 13:51

On the left, $\lfloor\arcsin(1-\epsilon)\rfloor=\lfloor\frac\pi2-\delta\rfloor=1$. The function is indeed undefined on the right but this doesn't matter as a limit is to be computed in the domain only. Thus,

$$\lim_{x\to 1}\lfloor\arcsin x\rfloor=\lim_{x\to 1^-}\lfloor\arcsin x\rfloor=1.$$