# Help with this multivariable limit involving the exponential [closed]

Can anyone give me some hints on how to solve this limit: $$\lim_{(x,y)\to (0,0)} \frac{e^x-e^y}{x^2y^2}$$ I have considered using power series of e or changing to polor coordinates but neither have yielded any results that can prove whether this limit exists or not. Any hints or suggestions would be helpful. Thank you.

• A usual tactic to tackle these limits is to check the limit along lines of the form $y=mx$. – player3236 Oct 21 at 15:16

We have that for $$x=y=t$$

$$\frac{e^x-e^y}{x^2y^2}=\frac{e^t-e^t}{t^4}=0$$

and for $$x=t$$, $$y=-t$$ with $$t \to 0^+$$

$$\frac{e^x-e^y}{x^2y^2}=\frac{e^t-e^{-t}}{t^4}=\frac{2}{t^3e^t}\frac{e^{2t}-1}{2t} \to \infty$$

• Sorry, my mistake. – Andrei Oct 21 at 15:25
• @strawberry-sunshine I'm assuming $t\to 0^+$ and since in this case the expression tends to $\infty$ this suffices to conlcude that the limit for the given function doesn't exist. Of course we can only use the second case to conclude considering the limit for $1/t^3$ as $t\to 0$. – user Oct 21 at 15:29
• Yes, I agree. Sorry, I missed $t \to 0^+$ earlier. – strawberry-sunshine Oct 21 at 15:32
• @strawberry-sunshine That's fine, your idea is indeed very good and smart! – user Oct 21 at 15:50

Consider the line $$y = -x$$, along which the limit becomes $$\lim_{x \to 0} \frac{e^x - e^{-x}}{x^4}$$ which clearly does not exist. $$\lim_{x \to 0^+} \frac{e^x - e^{-x}}{x^4} = \infty \text{ and }\lim_{x \to 0^-} \frac{e^x - e^{-x}}{x^4} = -\infty$$You may try to show this using the Maclaurin expansion of $$e^x$$, or by some other way.

If the original limit did exist, it would exist for $$y = -x$$ too.