# Multivariable Limit Question-Arctan and ln

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 !

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

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