Limit of 2 variable function

Question

I'm trying to determine the limit of this function: $$\lim_{(x,y)\to (1,2)} \frac{xy^2-4xy-y^2+4x+4y-4}{x^2+y^2-2x-4y+5}$$ I tried to approach in many different ways, such as $$\lim_{t\to 1} f(t,2t) \quad, \quad\lim_{t\to 1} f(t,2) \quad, \quad \lim_{t\to 2} f(1,t) $$ But i got that the limit is 0 for all of them, tried with polar coordinates but it seems hopeless to get the limit! How should I think there? Thanks in advance!

Answer 1

Let $f(x,y)=xy^2-4xy-y^2+4x+4y-4$ and $g(x,y)=x^2+y^2-2x-4y+5$. Then, for each $(x,y)\in\Bbb R^2$,\begin{align}f(x,y)&=f\bigl((x-1)+1,(y-2)+2\bigr)\\&=(x-1)(y-2)^2\end{align}and\begin{align}g(x,y)&=g\bigl((x-1)+1,(y-2)+2\bigr)\\&=(x-1)^2+(y-2)^2.\end{align}Therefore\begin{align}\lim_{(x,y)\to(1,2)}\frac{xy^2-4xy-y^2+4x+4y-4}{x^2+y^2-2x-4y+5}&=\lim_{(x,y)\to(1,2)}\frac{f(x,y)}{g(x,y)}\\&=\lim_{(x,y)\to(1,2)}\frac{(x-1)(y-2)^2}{(x-1)^2+(y-2)^2}\\&=\lim_{(x,y)\to(1,2)}(x-1)\overbrace{\frac{(y-2)^2}{(x-1)^2+(y-2)^2}}^{\phantom{[0,1]}\in[0,1]}\\&=0.\end{align}