# How to prove that $\frac{1}{x^4+y^4} e^{-\frac{1}{x^2+y^2}}\to0$ when $(x,y)\to (0,0)$?

How can I calculate the limit $$\lim_{(x,y)\to (0,0)} \frac{1}{x^4+y^4} e^{-\frac{1}{x^2+y^2}}$$ and show that it is zero?

When switching to polar coordinates, I get: $$\lim_{(x,y)\to (0,0)} \frac{1}{r^4 (\cos^4 \theta+\sin^4 \theta) } e^{-\frac{1}{r^2}}$$ but I have no idea how to show the term $\frac{1}{r^4 (\cos^4 \theta+\sin^4 \theta) }$ is bounded.

• Would it be valid to determine the limit of the function as $y$ approaches $x,$ and when $x$ approaches $0$? – K. Jiang May 17 '16 at 13:34
• no. because there are instances where the limit on one path is some constant, but on another path the limit is another constant, so the limit does not exist – yehushua May 17 '16 at 13:40
• Try to use the fact that $\cos^4\theta+\sin^4\theta$ is never zero. – Eclipse Sun May 17 '16 at 13:41
• An expression for $\cos^4\theta+\sin^4\theta$ is also given in this post: Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$. – Martin Sleziak Feb 16 '17 at 15:41

You need to show that $\cos^4\theta+\sin^4\theta$ never gets too small. A nice way to do this is to show

\begin{align} \cos^4\theta+\sin^4\theta &=(\cos^2\theta+\sin^2\theta)^2-2\cos^2\theta\sin^2\theta\\ &=1-{1\over2}(2\sin\theta\cos\theta)^2\\ &=1-{1\over2}\sin^22\theta\\ &\ge1-{1\over2}\\ &={1\over2} \end{align}

• @yehushua, I'm glad it helped. – Barry Cipra May 17 '16 at 18:34

To bound the trigonometric part: $$\cos^4 \theta + \sin^4 \theta = \cdots = \frac{\cos 4t + 3}{4}$$ So this factor lies between $1/2$ and $1$ (See Barry Cipra's answer to derive this lower bound).

It then suffices to show that: $$\lim_{r \to 0} \frac{e^{-1/r^2}}{r^4}= 0$$

Another way of getting Barry Cipra's result:

$c^4+s^4 =c^4+(1-c^2)^2 =c^4+1-2c^2+c^4 =1+2c^4-2c^2 =1-2c^2(1-c^2) \ge 1/2$ since $x(1-x) \le 1/4$ for $0\le x \le 1$.