How can I prove this multivariable function is continuous at (0,0)?

I have a function $$g(x,y) = \begin{cases} \frac{e^x-e^y}{x-y} & \text{if} & x \neq y \\ e^x & \text{if} & x = y\end{cases}$$

How can I show that this function is continuous at $(0,0)$ (and also at all $(x,y)$, but I believe this will be easy to generalise if I can do it for $(0,0)$.

I have tried to do it by definition, showing that $|\frac{e^x-e^y}{x-y}-e^0| < \varepsilon$ for $||(x,y)|| < \delta$, but I somehow can't find a way to do it. I have also tried to convert to polar coordinates but it doesn't seem to help.

Solution 1:

Let's use $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots +y^{n-1})$: $$\frac{e^x-e^y}{x-y}=\frac{1}{x-y}\left(\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}- \sum\limits_{n=0}^{\infty}\frac{y^n}{n!}\right)=1+\sum\limits_{n=2}^{\infty}\frac{x^{n-1}+x^{n-2}y+\cdots +y^{n-1}}{n!}$$ Last series converged and we need to prove, that it limit is $0$, when $(x,y)\to(0,0)$. Having $|x|<\delta$ and $|y|<\delta$ we can write $$\left|\sum\limits_{}^{}\frac{x^{n-1}+x^{n-2}y+\cdots +y^{n-1}}{n!} \right |\leqslant \sum\limits_{}^{}\frac{n\delta^{n-1}}{n!}=\sum\limits_{}^{}\frac{\delta^{n-1}}{(n-1)!}$$ So, last series is estimated by $e^\delta-1\to 0$, when $ \delta \to 0$.