I need a hint on what kind of path I should choose to show that this limit does not exist:
$\displaystyle\lim_{(x,y) \to (0,2)}\frac{(xy - 2x)}{x^3 + (y - 2)^2}$
I need a hint on what kind of path I should choose to show that this limit does not exist:
$\displaystyle\lim_{(x,y) \to (0,2)}\frac{(xy - 2x)}{x^3 + (y - 2)^2}$
We have \begin{equation} \begin{split} \displaystyle\lim_{(x,y) \to (0,2)}\frac{(xy - 2x)}{x^3 + (y - 2)^2} & = \displaystyle\lim_{(x,y) \to (0,2)}\frac{x(y - 2)}{x^3 + (y - 2)^2} \end{split} \end{equation}
If $f(x,y)=\frac{xy)}{x^3 + y^2}$ and $\varphi(x,y)=(x,y-2)$ then $ \displaystyle\lim_{(x,y)\to(0,+2)}\varphi(x,y)=(0,0) $ and $ \displaystyle\lim_{(x,y)\to(0,0)}f(\varphi(x,y))= \displaystyle\lim_{(x,y) \to (0,2)}\frac{x(y - 2)}{x^3 + (y - 2)^2} $ But $\displaystyle\lim_{(x,y)\to(0,0)}f(x,y)$ don't exist, by cause if $x=y=t$ we have $ \displaystyle\lim_{(x,y) \to (0,0)}\frac{xy}{x^3 + y^2} = \displaystyle\lim_{t \to 0}\frac{t^2}{t^3 + t^2} = \displaystyle\lim_{t \to 0}\frac{1}{t + 1}=1 $ and if $x=t$, $y=2t$ we have $ \displaystyle\lim_{(x,y) \to (0,0)}\frac{xy}{x^3 + y^2} = \displaystyle\lim_{t \to 0}\frac{2t^2}{t^3 + 4t^2} = \displaystyle\lim_{t \to 0}\frac{2}{t + 4}=\frac{1}{2} $