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Here is my limit:

$$ \lim \limits_{x,y \to 0,0}{(1 + x^2 y^2)}^{-\frac{1}{x^2 + y^2}}$$

I have learned two methods. One where we replace y with for example $y = kx $ (because $y = y_0 + k(x - x_0)$ and $y_0 = 0, x_0 = 0$). Or with $x = r *cos(\phi)$ and $x = r *sin(\phi)$ where $r \to 0$.

Neither seem to help me at the moment (or at least when I tried solving with both I didn't get a good answer.

It kind of seems like I could use $ \lim \limits_{x \to \infty}{(1 + \frac{1}{x})}^{x} = e$, but I tried and also couldn't get a decent answer.

Any ideas?

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We can use your idea of setting $x= r \cos \theta$, $y=r \sin \theta$ and the limit becomes

$$\lim_{r \rightarrow 0} \left(1+\frac{r^4 \sin^2 2\theta}{4} \right)^{-\frac{1}{r^2}}$$

For a given $r$, the maximum and minimum values of this function are $1$ and $\left( 1+\frac{r^4}{4} \right)^{-\frac{1}{r^2}}$ obtained by setting $\theta =0$ and $\theta = \frac{\pi}{4}$ respectively.

The second limit as $r\rightarrow 0$ is $\lim_{r\rightarrow 0}\left( 1+\frac{r^4}{4} \right)^{-\frac{1}{r^2}} = \lim_{x\rightarrow \infty} \left( 1+\frac{1}{x^2} \right)^{\frac{x}{2}} = 1$

where we have taken $x = \frac{2}{r^2}$

Because this the minimum value the function can take on the circle, we can say the following:

For any $\epsilon >0$ there exists $r$ such that $x^2 + y^2 < r^2 \Rightarrow |(1+x^2y^2)^{-\frac{1}{x^2+y^2}} -1| < \epsilon$ so the limit is $1$.

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  • $\begingroup$ WolframAlpha gives me the answer 1 somehow: wolframalpha.com/input/… Any idea what that is about ? $\endgroup$ – PadaKatel Mar 29 '16 at 13:31
  • $\begingroup$ I think you forgot the squares in $x^2y^2$ when switching to polar. $\endgroup$ – StackTD Mar 29 '16 at 13:36
  • $\begingroup$ Every computer algebra system has its limits (no pun intended), especially when it comes to evaluating multivariable limits. $\endgroup$ – tilper Mar 29 '16 at 13:37
  • $\begingroup$ The limit exists and is equal to $1$. You made a mistake evaluating $x^2y^2 = r^2 \sin 2 \theta$. Actually it is $x^2y^2 = r^4 \sin^2 2 \theta$. $\endgroup$ – Crostul Mar 29 '16 at 13:38
  • $\begingroup$ Thanks for the comments. Missed that the first time around. Apologies @PadaKatel for the massive mistake. $\endgroup$ – hexomino Mar 29 '16 at 13:47

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