Does $\sin^{-1}x$ has a vertical tangent I read that the function $f(x)$ has a vertical tangent at $x=a$ in the domain of $f$ if
$$f'(a^-) \to +\infty$$ and $$f'(a^+) \to +\infty$$ Or both approach to $-\infty$.
But for $f(x)=\sin^{-1}x$ $f'(x)=\frac{1}{\sqrt{1-x^2}}$ and we have
$$f'(-1^+) \to +\infty$$ and $$f'(1^-) \to +\infty$$ but $f'(-1^-)$ and $f'(1+)$ are not defined. Can we still say that  it has Vertical tangent at $x=\pm 1$
 A: The definition needs to allow for $+\infty$ on one side and $-\infty$ on the other, for example consider $f(x)=x^{2/3}$ which does have a vertical tangent at $x=0,$ yet the derivative $f'(x)=(2/3)x^{-1/3}$ goes to $+\infty$ as $x \to 0^+$ and to $-\infty$ as $x \to 0^-.$
Also, for a function at a domain endpoint, such as arcsin at $x=1,$ the limit of $\pm \infty$ for the derivative is only required for the approach from the side which stays in the domain of the function. 
The same thing happens in the definition of limit at a domain endpoint. I've seen some calc books which are sloppy about this, and insist that the limits from the left and from the right need to exist at $a$ in order that the limit at $a$ should exist. However even for $f(x)=\sqrt{x},$ which we certainly want to say has the limit $0$ at $x=0,$ the two-sided limit would fail to exist. In more complete definitions, a limit for a function with domain $D$ is still said to exist as $x \to c,$ provided that as $x\to c$ through points in $D$ we have $f(x) \to L$ for some real $L.$ For this it is only required that there be at least one sequence of points in $D$ which approaches $c,$ where during the approach the $x$ in $D$ are only restricted by $x \neq c.$ This works well for example if the function is only defined on the rationals, and one wants a limit as $x$ approaches some irrational.
