<s>[INST] 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.
 [/INST] 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*}
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