<s>[INST] Find $\lim\limits_{(x,y)\to(1,-1)}{\frac{e^{x-y}\tan{(x+y)}}{x^2-y^2}}$ I need to find $\lim\limits_{(x,y)\to(1,-1)}{\frac{e^{x-y}\tan{(x+y)}}{x^2-y^2}}$. I came up with this:
Since $x^2-y^2$ changes its sign, I can'y apply $\tan{(x+y)}\sim(x+y)$ just yet. So,$\lim\limits_{(x,y)\to(1,-1)}{|\frac{e^{x-y}\tan{(x+y)}}{x^2-y^2}|}=\lim\limits_{(x,y)\to(1,-1)}{\frac{e^{x-y}|\tan{(x+y)}|}{|x^2-y^2|}}=\lim\limits_{(x,y)\to(1,-1)}{\frac{e^{x-y}|x+y|}{|x-y|.|x+y|}}=\lim\limits_{(x,y)\to(1,-1)}{\frac{e^{x-y}}{|x-y|}}=\frac{e^2}{2}$ 
Is this correct and is there a better way to approach the problem?
 [/INST] What you did was to compute the limit of $\left|\frac{e^{x-y}\tan(x+y)}{x^2-y^2}\right|$, not of $\frac{e^{x-y}\tan(x+y)}{x^2-y^2}$.
You have$$\lim_{(x,y)\to(1,-1)}\frac{e^{x-y}\tan(x+y)}{x^2-y^2}=\lim_{(x,y)\to(1,-1)}\frac{e^{x-y}}{x-y}\frac{\tan(x+y)}{x+y}=\frac{e^2}2,$$since$$\lim_{(x,y)\to(1,-1)}\frac{\tan(x+y)}{x+y}=\lim_{h\to0}\frac{\tan h}h=\tan'(0)=1.$$
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