Differentiability for arcsin and limits[Question]

Question

So continuing from the title; I have a question regarding the arcsin. In my book it says that the inverse of sin is differentiable only on the open interval $(-1,1)$ Which I can understand.

However, later they claim that the slope of the graph approaches infinity when $x->-1+$ or as $x->1-$

And following the formula for the derivative for arcsin

$\frac{d}{dx}arcsin=\frac{1}{sqrt{1-x^2}}$

I can see that. Although, would't it mean there has to be a asymptote at $x=-1$ and $x=1$ if the slope appraoches infinity at $x->-1+$ or as $x->1-$?

Thank you in advance!

Answer 1

If you look at the graph, you'll notice that $\arcsin(x)$ is bounded on $[-1,1]$, so there can't be an asymptote.

Asymptotes occur when the function approaches infinity. Here, the derivative of the function approaches infinity. A similar thing happens with $\sqrt[3]{x}$ at $x = 0$. Does this look like it has an asymptote?

Answer 2

Yes the tangent line at that points is vertical but there are not asymptotes.

It’s analogue at the situation for $f(x)=\sqrt x$ at x=0.