# Multivariable limit $\lim_{(x,y)\to(0,0)} (x^2+y^2)^{x^2y^2}$

I'm studying multivariable limits and I have a problem regarding this limit:

$$\lim_{(x,y)\to(0,0)} (x^2+y^2)^{x^2y^2}.$$ I've found that it is equal to 1, by rewriting the limit using $t = e^{ln(t)}$. Despite this, the answer in the book is 0. What is the correct answer?

Using polar coordinates, we get

$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\implies (x^2+y^2)^{x^2y^2}=r^{2r^4sin^2t\cos^2t}=r^{\frac12r^4\sin^22t}=e^{\frac12r^4\sin^22t\log r}\xrightarrow[r\to 0^+]{}e^0=1$$

so I'd go with you being correct and the book is wrong. What book is that, anyway?

• Shouldn't it be $r^4$ in the exponent? – Julián Aguirre Nov 14 '14 at 12:11
• It's a college book from Romania :). Why it is correct to make that substitution? The limit means that we should consider all the pats to (0,0) not only those whose points lie on a straight line. – SebiSebi Nov 14 '14 at 12:13
• @timbuc consider Henrik's answer also – godonichia Nov 14 '14 at 12:20
• First, the above doesn't consider only straight paths, as the limit remains whatever the angle $\;t\;$ is. Second, it shows the limit cannot be zero as at least in the above way we got something different. La revedere acum. – Timbuc Nov 14 '14 at 12:21
• Thank you, Julian. Editing – Timbuc Nov 14 '14 at 12:21

If we examine what happens when we approach $(0,0)$ along the line $x=0$ (or $y=0$), we get: $$\lim_{(x,y)\to(0,0)} (x²+y²)^{x²y²}= \lim_{y\to0} (0+y^2)^{0}= 1$$

meaning that if the limit exists it must be $1$.

• I'll vote up this answer as it is seriously shorter, simpler and more elegant that any other here, including mine. +1 – Timbuc Nov 14 '14 at 12:28
• This is incorrect: for example this limit $$\lim_{(x,y)\to(0,0)}\frac xy$$ doesn't exist however if you pass first to the limit $x\to0$ then you'll find the limit $0$. – user63181 Nov 14 '14 at 12:33
• That doesn't really change anything. I wrote "if the limit exists", I didn't claim it does. Applying the same technique to your example the conclusion would be that if the limit exists (which it doesn't) it must be 0. The point is simply the SebiSebi's book is wrong, the limit can't be 0. – Henrik Nov 14 '14 at 12:47
• And in case anyone wonders why the limit in @SamiBenRomdhane's comment doesn't exist, just consider what happens if we approach $(0,0)$ along the line $x=y$? – Henrik Nov 14 '14 at 12:50
• @Timbuc WA is utter sh*t when it comes to multivariable limits. I think what it does is that it computes the iterated limits. As for the first equality in this answer, it's not problematic at all if you start with the assumption that the limit exists. Then the equality is perfectly legal and allows us to conclude that the limit can't be zero. – Git Gud Nov 14 '14 at 13:12

Let $x=r\cos t$ and $y=r\sin t$ so

$$\lim_{(x,y)\to(0,0)} (x^2+y^2)^{x^2y^2}=\lim_{r\to0}r^{\frac12r^4\sin^2(2t)}=1$$ because

$$\lim_{r\to0}r\ln r=0$$

• Please check your solution – godonichia Nov 14 '14 at 12:25
• @SamiBenRomdhane, perhaps the same mine had: it must be $\;r^4\;$ in the exponent. – Timbuc Nov 14 '14 at 12:27
• Yes thanks! Fortunately, it doesn't change the result:-) – user63181 Nov 14 '14 at 12:29
• It is the same thing I did, so I thought the limit was indeed $1$. But wolfram tells us is $0$! wolframalpha.com/input/… – Ant Nov 14 '14 at 15:25
• @Ant graph the function $f(x,y)=(x^2+y^2)^{x^2y^2}$ and see what happens around $(x,y)=(0,0)$... :). It's so strange, even in 2018 Wolfram says that the limit is $0$! – manooooh Sep 19 '18 at 4:34

The limit cannot be $0$ because if so, then

$$\sqrt{x^{2} + y^{2}} < \delta \Rightarrow |(x^{2} + y^{2})^{(xy)^{2}}| < \epsilon,$$ $$x^{2} + y^{2} < \delta^{2}; x^{2}, y^{2} < \delta^{2},$$ $$(x^{2} + y^{2})^{(xy)^{2}} < \delta^{2\delta^{4}} \leq \epsilon,$$ $$\delta \ln \delta = (\frac{\ln \epsilon}{2})^{1/4}.$$ If $$\epsilon := 1/2$$ then $(\ln \epsilon /2)^{1/4}$ becomes meaningless.