Behavior of arccos(x) near x=-1

Question

Actual problem is

"Is $\theta=\pi$ valid for polar coordinate representation of parabola"

Where it's equatioin is $\theta(r)=arccos(\frac{a}{r}-1)$ a is some positive constant

My friend says if $ r \to \infty$ then it will yield $\theta (r=\infty)=\pi$

but i don't think so(also, for i've seen several references of parabola omitting r goes infinity or $\theta=(2n-1)\pi$ or $2n\pi$ where n is integer)

Thus I want to see arccosine's behavior around -1 so that i'll show that

$\lim\limits_{r \to \infty} \frac{1}{r}$ is not small enough to make $\lim\limits_{r \to \infty}arccos(\frac{a}{r} -1)=\pi$

I've thought about taylor expanding arccos(x) around x=-1

and i have found

arccos(-1+x)=$\pi -\frac{1}{\sqrt {1-z^2}}x -\frac{z}{(1-z^2)^{3/2}}\frac{x^2}{2}$....... where z=-1 x:positive and very small

it seems that every coefficients explode for z=-1 and i see that even every derivatives of coefficients explode when z=-1

so i think every coefficients are far greatly bigger term even multiplied with very small x which $\lim x \to 0$

Therefore I think inverse cosine is too sensitive to make $\lim\limits_{r \to \infty}arccos(\frac{a}{r}-1)=\pi$(or should i say sum doesn't converge however x is small in polynomial sense?)

I'm bit into math, and know little bit of analysis but i haven't studied analysis deeply

1. is $\theta=\pi$ valid for parabola (or $2n\pi$ but i'm sure you know what i mean) r=$\frac{a}{1+cos\theta}$ <--primary question

2. want to know $\lim\limits_{r \to \infty}arccos(\frac{a}{r}-1)=\pi$ even if arccosine is super sensitive or whatever

Answer 1

The polar equation

$$r(\cos\theta +1)=a$$ indeed represents parabolas, and close to $\theta =\pi$, you can write

$$a=r(\cos(\pi+\delta) +1)=r(-\cos\delta+1)\approx \frac{r\delta^2}2.$$

When $r$ tends to infinity, $\delta$ to zero and $\theta$ to $\pi$ (as fast as $r^{-1/2}$). There is no mystery here.

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Q1. $\theta=\pi$ is of course not allowed, as $r\cdot0=a$ has no solution.

Q2. By continuity of the arc cosine,

$$\lim_{r\to\infty}\arccos\left(\frac ar-1\right)=\arccos\left(\lim_{r\to\infty}\frac ar-1\right)=\pi.$$

Note that the Taylor expansion of the arc cosine around $-1$ does not exist.

Answer 2

Yes the limit will go to pi. It's not about sensitivity. In fact a function does not even have to be defined at the limit. The definition is for all epsilom > 0 there exists a finite r that gives an angle in pi +/- epsilom