Limit of a two variable function involving arcsin and arctan

Question

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.

Answer 1

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

$$\lim_{(x,y)\to (0,0)}\frac{\arcsin(x+2y)}{\arctan(2x+4y)}=\frac12$$

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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 $0<|t|=|x+2y|<\delta$.

Answer 2

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*}

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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.