Question: how can i compute this limit without using Taylor series?

$\lim\limits\_{(x,y) \to(0,0)} \frac{e^{x^2-y^2} -1}{ x - y}$, while $x\neq y$.

i tried to put $r^2=x^2-y^2$ and transfer to polar but it didn't work for me.

Thank you for your help.

Answer: 
Define a function $f$ on $\mathbb{R}$ by $f(u)=(\exp(u)-1)/u$ if $u\not =0$, and $f(0)=1$. Then $f$ is continuous on $\mathbb{R}$(to show that, we need only use that the derivative of $\exp(u)$ at $u=0$ is $1$). We have $uf(u)=\exp(u)-1$ for all $u\in \mathbb{R}$. Let $g$ your function: we have
$$g(x,y)=\frac{(x^2-y^2)f(x^2-y^2)}{x-y}=(x+y)f(x^2-y^2)$$
Hence $g(x,y)$ is continuous on $\mathbb{R}^2$ as the product of two continuous functions, it is easy to finish: the limit is $0\times 1=0$.
By making the substitution $z = e^{x^2-y^2}-1$, we obtain that
\begin{align\*}
\lim\_{(x, y)\rightarrow (0, 0)}\frac{e^{x^2-y^2}-1}{x^2-y^2} &= \lim\_{z\rightarrow 0}\frac{z}{\ln(1+z)}\\
&= \lim\_{z\rightarrow 0}\frac{1}{\ln\left((1+z)^{1/z}\right)}\\
&=1.
\end{align\*}
Then,
\begin{align\*}
\lim\_{(x, y)\rightarrow (0, 0)}\frac{e^{x^2-y^2}-1}{x-y} &= \lim\_{(x, y)\rightarrow (0, 0)}\frac{e^{x^2-y^2}-1}{x^2-y^2} \frac{x^2-y^2}{x-y}\\
&=\lim\_{(x, y)\rightarrow (0, 0)}\frac{e^{x^2-y^2}-1}{x^2-y^2} (x+y)\\
&=\lim\_{(x, y)\rightarrow (0, 0)}\frac{e^{x^2-y^2}-1}{x^2-y^2} \lim\_{(x, y)\rightarrow (0, 0)}(x+y)\\
&=0.
\end{align\*}