I need to prove that the limit does not exits

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

I tried finding the limit as f(x) approaches from x and y axes but I still get an indeterminate answer.

  • $\begingroup$ something error showing in your function $\endgroup$
    – Groups
    Oct 12, 2015 at 11:10
  • $\begingroup$ As you approach along the $x$ or $y$ axes, the function is identically zero (the numerator vanishes), so even though you get an indeterminate of $\frac{0}{0}$, the limit is $0$. $\endgroup$ Oct 12, 2015 at 11:14
  • 4
    $\begingroup$ What about trying other trajectories to approach $(0,0)$? For instance, $y=x$, $y=2x$... $\endgroup$
    – Miguel
    Oct 12, 2015 at 11:14

1 Answer 1


When $y=0$ and $x\not=0$, $$ f(x,y)=f(x,0)=\frac{e^{x\cdot 0}-1}{x^2+0^2}=\frac{e^0-1}{x^2}=\frac{1-1}{x^2}=\frac{0}{x^2}=0. $$ Therefore, whenever $x\not=0$, and $(x,y)$ is on the $x$-axis, this function is identically zero. Therefore, the limit as $x$ approaches zero is $0$.

Now, if you try the trajectory where $x=y$, then $$ f(x,y)=f(x,x)=\frac{e^{x^2}-1}{2x^2}. $$ As $x$ (and $y$) approach zero, you do get an indeterminate form, but a single application of l'Hopital's rule (or looking at the power series expansion) will show that this limit is not zero.

  • $\begingroup$ wouldn't l'hopital's rule turn the limit into 2xe^(x^2)/4x.Then applying it again would give you 4x^2(e^(x^2))/4 making the limit 0? $\endgroup$
    – michael
    Oct 12, 2015 at 11:26
  • $\begingroup$ @michael Cancel your $x$'s to get $\frac{1}{2}e^{x^2}$. Also, it looks like your second application of l'Hopital's rule forgot the product rule in the numerator. $\endgroup$ Oct 12, 2015 at 12:52
  • $\begingroup$ More precisely, the derivative of $2xe^{x^2}$ is $2e^{x^2}+4x^2e^{x^2}$. $\endgroup$ Oct 12, 2015 at 13:16

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