existence of a limit with trig functions for a very long time now i've been going nuts trying to track down and/or produce on my own a proof of the following, which i'm empirically certain is true. suppose that 
$$
(-1) \leqslant k \leqslant (+1) 
$$$$
f(x,y) = \arccos[(\cos x)(\cos y) + k(\sin x)(\sin y)]
$$$$
g(x,y) = (\sin^2 x) - 2k(\sin x)(\sin y) + (\sin^2 y)
$$$$
h(x,y) = f(x,y)^2 / g(x,y)
$$
what i want to show is that $h(x,y)$ has a limit (namely $1$) as $(x,y)$ tends to $(0,0)$. please help if you have any insights about this. i'm so tired of being defeated by it.
thanks if you can help
peace
stm
 A: Try Taylor series.  So $\cos(x)\cos(y) + k \sin(x) \sin(y) = 1 + kxy -x^2/2 - y^2/2 + \text{h.o.t}$.  Now since $\cos(x) = 1 - x^2/2 + \text{h.o.t}$, we see that $\arccos(1 - x^2/2) = x + \text{h.o.t}$, or $\arccos(1-z) = \sqrt{2z} + \text{h.o.t}$.  So
$$ \arccos(\cos(x)\cos(y) + k \sin(x) \sin(y) = \sqrt{x^2 + y^2 - 2kxy} + \text{h.o.t} .$$
Do a similar thing with $g(x,y)$ (which will be much easier), and the result follows.
Here $\text{h.o.t}$ means "higher order terms."
A: This is a quite tricky problem, since we need an approximation of $\arccos$ near $1$, where it is not analytic. I shall assume $|k|<1$, since $g(x,\pm x)=0$ when $k=\pm1$.
When $0\leq t\leq1$ then $\phi:=\arccos t\in\bigl[0,{\pi\over2}\bigr]$ and $\sin\phi=\sqrt{1-t^2}$. It follows that
$$\arccos t=\phi=\arcsin\sqrt{1-t^2}\qquad(0\leq t\leq1)\ ,$$
from which we infer that for $0\leq u\leq 1$ one has
$$\arccos(1-u)=\arcsin\sqrt{2u-u^2}=\sqrt{2u}\ \sqrt{1-u/2}\ {\arcsin\sqrt{2u-u^2}\over\sqrt{2u-u^2}}\ .$$
This implies
$$\lim_{u\to0+}{\arccos^2(1-u)\over 2u}=1\ .\tag{1}$$
In our case
$$\eqalign{u=u(x,y)&:=1-\cos x\cos y -k\sin x\sin y\cr
&\>\, ={1-k\over2}\bigl(1-\cos(x+y)\bigr)+{1+k\over2}\bigl(1-\cos(x-y)\bigr)\cr &\>\,= (1-k)\sin^2\alpha +(1+k)\sin^2\beta\ ,\cr}$$
where we have put $\alpha:={x+y\over2}$, $\>\beta:={x-y\over2}$. It follows that $$u(x,y)>0\quad\bigl((x,y)\ne(0,0)\bigr),\qquad\lim_{(x,y)\to(0,0)} u(x,y)=0\ .$$
On the other hand
$$\eqalign{g(x,y)&={1-k\over2}(\sin x+\sin y)^2+{1+k\over2}(\sin x-\sin y)^2\cr
&=2(1-k)\sin^2\alpha\cos^2\beta+2(1+k)\cos^2\alpha\sin^2\beta\cr
&=2u(x,y)-4\sin^2\alpha\sin^2\beta\ .\cr}$$
Therefore we obtain
$${g(x,y)\over 2u(x,y)}=1-{2\sin^2\alpha\sin^2\beta\over(1-k)\sin^2\alpha +(1+k)\sin^2\beta}\to1\qquad\bigl((x,y)\to(0,0)\bigr)\ ,
$$
which together with $(1)$ proves the claim.
