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Can someone help me calculate : $$\lim _{(x,y)\to (1,2)} \frac {\arctan(x+y-3)}{\ln(x+y-2)}?$$

I think substituting $x+y = t $ might help, but I am not sure that doing such a substitution in a multivariable case is legitimate, and I prefer not doing this.

Can you help me?

Thanks !

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l'Hôpital's rule? –  Douglas B. Staple Apr 16 '13 at 15:33

2 Answers 2

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It is better to take $t=(x-1)+(y-2)$. Then we obtain $$ \lim_{t\to 0}\frac{\arctan{t}}{\ln(1+t)}=1. $$

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$$\lim\limits_{(x,y)\to (1,2)} \frac {\arctan(x+y-3)}{\ln(x+y-2)}=\lim\limits_{(x,y)\to (1,2)} \frac {\arctan((x-1)+(y-2))}{\ln((x-1)+(y-2))}=\\ =\left|\matrix{u=x-1\\ v=y-2} \right|=\lim\limits_{(u,\,v)\to (0,0)} \frac {\arctan(u+v)}{\ln(1+u+v)}=1$$ since both numerator and denominator depends only on sum $u+v.$

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Thanks !!!!!!!!!!!!!! –  czash Apr 16 '13 at 15:53

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