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The question is about finding
$$lim_xto1 f(x)$$
where
$$f(x) = lfloorsin^-1(x)rfloor$$

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 $arcsin(1) = fracpi2$, 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 $mathrmfloor(arcsin(x))$ 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.

You are correct, the limit is only defined as:

$$lim_xto 1^- bigllfloor arcsin (x)bigrrfloor$$

and since for $xto 1^-$ we have $arcsin (x)to fracpi2 gt1$, we have that the limit is $1$.

On the left, $lfloorarcsin(1-epsilon)rfloor=lfloorfracpi2-deltarfloor=1$. 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,

$$lim_xto 1lfloorarcsin xrfloor=lim_xto 1^-lfloorarcsin xrfloor=1.$$