computing $\lim\limits_{(x,y) \to(0,0)} \frac{e^{x^2-y^2} -1}{ x - y}$ without Taylor series 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.
 A: 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 $0\times 1=0$.
A: (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. 
A: 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*}
