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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?

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2 Answers 2

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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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  • $\begingroup$ Thank you very much! $\endgroup$ Jul 7, 2021 at 7:31
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We have $$\frac{e^{x-y}\tan(x+y)}{x^2-y^2}=\frac{e^{x-y}}{x-y}\cdot\frac{\sin(x+y)}{x+y}\cdot\frac{1}{\cos(x+y)}$$ Break this into three parts using composition of limits.

Part 1:

Set $z(x,y)=x-y$ so $z(x,y)\to 2$ as $(x,y)\to(1,-1)$. Then $$\lim\limits_{(x,y)\to(1,-1)}\frac{e^{x-y}}{x-y}=\lim\limits_{z\to 2}\frac{e^{z}}{z}=\frac{e^2}{2}$$

Part 2:

Set $z(x,y)=x+y$ so that $z(x,y)\to 0$ as $(x,y)\to(1,-1)$. Then $$\lim_{(x,y)\to(1,-1)}\frac{\sin(x+y)}{(x+y)}=\lim\limits_{z\to 0}\frac{\sin(z)}{z}=1$$

Part 3:

Set $z(x,y)=x+y$ again so that $z(x,y)\to 0$ as $(x,y)\to(1,-1)$. Then $$\lim\limits_{(x,y)\to(1,-1)}\frac{1}{\cos(x+y)}=\lim\limits_{z\to 0}\frac{1}{\cos(z)}=\frac{1}{\cos(0)}=1$$

Conclusion:

We have $$\lim\limits_{(x,y)\to(1,-1)}\frac{e^{x-y}\tan(x+y)}{x^2-y^2}=\frac{e^2}{2}\cdot 1\cdot 1=\frac{e^2}{2}$$

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    $\begingroup$ Thank you for your answer. $\endgroup$ Jul 7, 2021 at 7:32

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