limit {x,y -> 0,0} ( exp[-1/(x^2+y^2)] / (x^6+y^6) )

[jazzman]

Please help me understand how to calculate this limit:
\(\displaystyle \lim_{(x,y)\to(0,0)}\frac{e^{-\frac{1}{x^2+y^2}}}{x^6+y^6}\)

I have tried the following:

\(\displaystyle 0\le\frac{e^{-\frac{1}{x^2+y^2}}}{x^6+y^6}}\le\frac{e^{-\frac{1}{x^2}}}{x^6}\)

So I am left with:

\(\displaystyle \lim_{x\to0}\frac{e^{-\frac{1}{x^2}}}{x^6}\)

Which according to the calculator is equal 0 but I don't understand why.
I tried using L'hopital's rule with no success..
Perhaps taylor expansion will help?

Please help me understand!

 

[galactus]

Re: limit{x->0}(exp(-1/x^2)/x^6)

\(\displaystyle \displaystyle{\frac{e^{\frac{-1}{x^{2}}}}{x^{6}}=\frac{1}{x^{6}e^{\frac{1}{x^{2}}}}}\)

Now, can you see why it is 0?. The larger the denominator gets, the closer we get to 0. That's because as x gets closer to 0, then the e term becomes unbounded very quickly and we approach 0.

 

[jazzman]

Re: limit{x->0}(exp(-1/x^2)/x^6)

\(\displaystyle \displaystyle{\frac{e^{\frac{-1}{x^{2}}}}{x^{6}}=\frac{1}{x^{6}e^{\frac{1}{x^{2}}}}}\)

Now, can you see why it is 0?. The larger the denominator gets, the closer we get to 0. That's because as x gets closer to 0, then the e term becomes unbounded very quickly and we approach 0.

I see, you say because \(\displaystyle e^{-\frac{1}{x^2}}\) approaches zero faster than \(\displaystyle x^{6}\) then the limit is 0?
Is there a more formal proof?

 

[tkhunny]

Re: limit{x->0}(exp(-1/x^2)/x^6)

How formal would you like?

Substitute u = 1/(x^2) and ponder the limit as u increases without bound.