MHB How to Prove a Limit in Two Variables?

[alyafey22]

Hey MHB !

I've got a question that I am clueless how to proceed

Prove that

$$\Large \lim_{(x,y)\to (0,0)}(1+x^2y^2) ^{\frac{-1}{x^2+y^2}} = 1$$

Any hint would be appreciated.

 

[chisigma]

Hey MHB !

I've got a question that I am clueless how to proceed

Prove that

$$\Large \lim_{(x,y)\to (0,0)}(1+x^2y^2) ^{\frac{-1}{x^2+y^2}} = 1$$

Any hint would be appreciated.

Using polar coordinaters is...

$\displaystyle |\ln \{(1 + x^{2}\ y^{2})^{- \frac{1}{x^{2} + y^{2}}}\}| = \frac{\ln (1 + \rho^{4}\ \cos^{2} \theta\ \sin^{2} \theta)}{\rho^{2}} < \frac{\ln (1 + \rho^{4})}{\rho^{2}} = \frac{\rho^{4} - \frac{\rho^{8}}{2} + ...}{\rho^{2}}\ (1)$

... so that...

$\displaystyle \lim_{\rho \rightarrow 0} \ln \{(1 + x^{2}\ y^{2})^{- \frac{1}{x^{2} + y^{2}}}\} = 0\ (2)$

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

Have you considered converting to polar coordinates?

 

[alyafey22]

Using polar coordinaters is...

$\displaystyle |\ln \{(1 + x^{2}\ y^{2})^{- \frac{1}{x^{2} + y^{2}}}\}| = \frac{\ln (1 + \rho^{4}\ \cos^{2} \theta\ \sin^{2} \theta)}{\rho^{2}} < \frac{\ln (1 + \rho^{4})}{\rho^{2}} = \frac{\rho^{4} - \frac{\rho^{8}}{2} + ...}{\rho^{2}}\ (1)$

... so that...

$\displaystyle \lim_{\rho \rightarrow 0} \ln \{(1 + x^{2}\ y^{2})^{- \frac{1}{x^{2} + y^{2}}}\} = 0\ (2)$

Kind regards

$\chi$ $\sigma$

Nice trick of using logarithm !

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Have you considered converting to polar coordinates?

I did but with no success.

 

[MarkFL]

I did but with no success.

What I did was let:

$$k=\sin^2(\theta)\cos^2(\theta)$$

And then we have:

$$\lim_{r\to0}\left[\left(1+kr^4\right)^{-\frac{1}{r^2}}\right]=L$$

Taking natural logs:

$$-\lim_{r\to0}\left[\frac{\ln\left(1+kr^4\right)}{r^2}\right]=\ln(L)$$

Apply L'Hôpital:

$$-2k\lim_{r\to0}\left[\frac{r^2}{kr^4+1}\right]=\ln(L)$$

$$0=\ln(L)$$

$$L=1$$