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

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.

• something error showing in your function Oct 12, 2015 at 11:10
• 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 $0$. Oct 12, 2015 at 11:14
• What about trying other trajectories to approach $(0,0)$? For instance, $y=x$, $y=2x$... Oct 12, 2015 at 11:14

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.
• @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. Oct 12, 2015 at 12:52
• More precisely, the derivative of $2xe^{x^2}$ is $2e^{x^2}+4x^2e^{x^2}$. Oct 12, 2015 at 13:16