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: 
(Edit. The following answer is incorrect, the error is indicated in the comments.) 

I don't think the limit exists, as you obtain different values along different paths, as shown below. 

$\lim\limits\_{(x,y) \to(0,0)} \dfrac{e^{x^2-y^2} -1}{ x - y} = 
\lim\limits\_{(x,y) \to(0,0)} \dfrac{e^{(x+y)(x-y)} -e^{x-y}+e^{x-y} -1}{ x - y} =$  

$=\lim\limits\_{(x,y) \to(0,0)} \dfrac{e^{x-y}(e^{x+y} -1)}{ x - y} + 
\lim\limits\_{(x,y) \to(0,0)} \dfrac{e^{x-y} -1}{ x - y}$. 

Now $\lim\limits\_{(x,y) \to(0,0)} \dfrac{e^{x-y} -1}{ x - y}=1$ and 
$\lim\limits\_{(x,y) \to(0,0)} e^{x-y}=1$. But $\lim\limits\_{(x,y) \to(0,0)} \dfrac{(e^{x+y} -1)}{ x - y}$ does not exist, as it has different values along different paths, as shown below. 

For example, first look at the constraint $y=\frac12x$. Then $x-y=\frac12x$ and $x+y=\frac32x$. So $\lim\limits\_{(x,y) \to(0,0)} \dfrac{(e^{x+y} -1)}{ x - y} = \lim\limits\_{(x,y) \to(0,0)} \dfrac{(e^{\frac32x} -1)}{ \frac12x} = 
\lim\limits\_{(x,y) \to(0,0)} \dfrac{3(e^{\frac32x} -1)}{ \frac32x}=3$, along the path $y=\frac12x$. 

Now look at the constraint (or path) $y=\frac14x$. Then $x-y=\frac34x$ and $x+y=\frac54x$. So $\lim\limits\_{(x,y) \to(0,0)} \dfrac{(e^{x+y} -1)}{ x - y} = \lim\limits\_{(x,y) \to(0,0)} \dfrac{(e^{\frac54x} -1)}{ \frac34x} = 
\lim\limits\_{(x,y) \to(0,0)} \dfrac{5(e^{\frac54x} -1)}{ 3\cdot\frac54x}=\dfrac53\not=3$, along the path $y=\frac14x$. 

Since the value of the limit should be independent of the path along which the limit is evaluated, the above shows that the limit does not exist.
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\*}