Limit of a two variable function involving arcsin and arctan I recently came across this problem of finding the limit of a function in two variables as we approach the origin, defined as follows:
$$\lim_{(x,y)\to (0,0)} \frac{\arcsin(x+2y)}{\arctan(2x+4y)}$$
This was on the mid-term test on our course of Multivariable Calculus. While, all the paths that pass through the origin I have tested, seems to give the limit $\frac{1}{2}$, which seems reasonable through recognizing the fact that the argument of arctan is twice as that of arcsin and the logic of single variable limit: $\lim_{x\to 0} \frac{\arcsin(x)}{\arctan(2x)}$. It is, however, not sufficient to prove that the limit is indeed, $\frac{1}{2}$ since the two-path test is a test for the non-existence of limit and not the existence of one. I have even tried Sandwich Theorem and evaluating the limit by converting to polar coordinates but have reached to no conclusion, of course.
I am wondering if setting (x+2y) as some parameter t followed by tending t to zero would work, but then again that would be the same as testing along the path x=-2y.
I am truly dumbfounded by this problem and after long efforts at solving it, I have no answer. Any kind of hint towards which direction I should be proceeding or a solution would be highly appreciated.
 A: If one wishes to use the squeeze theorem, then we may proceed as follows.  Note that in This Answer, I showed that for $|x|\le 1$
$$|x|\le |\arcsin (x)|\le \frac{|x|}{\sqrt{1-x^2}} \tag 1$$
and in This Answer, I showed that 
$$\frac{|x|}{\sqrt{1+x^2}}\le |\arctan(x)|\le |x| \tag 2$$
Using $(1)$ and $(2)$, we see that 
$$\frac{x+2y}{2x+4y}\le \frac{\arcsin(x+2y)}{\arctan(2x+4y)}\le \frac{\frac{x+2y}{\sqrt{1-(x+2y)^2}}}{\frac{2x+4y}{\sqrt{1+(2x+4y)^2}}}$$
which simplifies to 
$$\frac12\le \frac{\arcsin(x+2y)}{\arctan(2x+4y)}\le \frac12 \sqrt{\frac{1+(2x+4y)^2}{1-(x+2y)^2}} \tag 3$$
Applying the squeeze theorem to $(3)$ we arrive at the coveted limit
$$\lim_{(x,y)\to (0,0)}\frac{\arcsin(x+2y)}{\arctan(2x+4y)}=\frac12$$

Alternatively, we may parameterize the problem by setting $t=x+2y$.  Then, as $(x,y)\to (0,0)$, $t\to 0$ also.  Note that this does not presume anything regarding a relationship between $x$ and $y$. We have merely defined a new variable as a linear combination of them.   All we need to know is that for all $\epsilon>0$, there exists a $\delta>0$ such that $$\left|\frac{\arcsin(t)}{\arctan(2t)}-\frac12\right|<\epsilon$$whenever $0<|t|=|x+2y|<\delta$.
A: First, we claim that 
$$
    \lim_{t\to 0} \frac{\arcsin{t}}{t} = 1
$$
This comes from L'Hôpital's rule, or if you like, the inverse function theorem (this limit is the definition of the derivative of arcsin at $t=0$).  
Given $\epsilon>0$, there exists a $\delta_1 > 0$ such that
$$ 0 < |t| < \delta_1 \implies \left|\frac{\arcsin t}{t} -1 \right| < \epsilon$$
Let $\delta = \frac{\delta_1}{3}$, and suppose $0<\sqrt{x^2 + y^2} < \delta$.  Then each of $|x|$ and $|y|$ are also $< \delta$, so
$$
   0 < |x+2y| \leq |x| + 2|y| < 3\delta = \delta_1
$$
Since $0 < |x+2y| < \delta_1$, we know
$$
    \left|\frac{\arcsin(x+2y)}{x+2y}-1\right| < \epsilon
$$
Since this works for any $\epsilon > 0$, we know
$$
    \lim_{(x,y) \to (0,0)} \frac{\arcsin(x+2y)}{x+2y} = 1
$$
In a very similar fashion, you can prove that
$$
    \lim_{(x,y) \to (0,0)} \frac{2x+4y}{\arctan(2x+4y)} = 1
$$
Putting these together,
\begin{align*}
    \lim_{(x,y) \to (0,0)} \frac{\arcsin(x+2y)}{\arctan(2x+4y)}
   &=\lim_{(x,y) \to (0,0)} \left(\frac{\arcsin(x+2y)}{x+2y}\cdot\frac{2x+4y}{\arctan(2x+4y)}\cdot\frac{1}{2}\right) \\
   &=\lim_{(x,y) \to (0,0)} \frac{\arcsin(x+2y)}{x+2y}
        \cdot \lim_{(x,y) \to (0,0)}\frac{2x+4y}{\arctan(2x+4y)}
        \cdot\frac{1}{2} \\
    &=1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}
\end{align*}

Come to think of it, you can probably start with
$$
  \lim_{t\to 0} \frac{\arcsin t}{\arctan 2t} = \frac{1}{2}
$$
and save a step.  But oh well.
