# What is the limit of lim_((x,y)->(0,0))x^3/(x^2+y^2)?

May 20, 2018

$\textcolor{b l u e}{{\lim}_{x \to 0 , \text{ "y=0} x^2/(x^2+y^2)!=lim_{y->0," } x = 0} {x}^{2} / \left({x}^{2} + {y}^{2}\right)}$
Thus, the original limit does not exist.

#### Explanation:

In order for this limit to exist, the fraction ${x}^{2} / \left({x}^{2} + {y}^{2}\right)$must approach the same value $L$

regardless of the path along which we approach$\left(0 , 0\right)$
Consider approaching$\left(0 , 0\right)$ along the x-axis That means fixing
$y = 0$ and finding the limit $\textcolor{red}{{\lim}_{x \to 0} {x}^{2} / \left({x}^{2} + {y}^{2}\right)}$

${\lim}_{x \to 0 , \text{ } y = 0} {x}^{2} / \left({x}^{2} + {y}^{2}\right) = {\lim}_{x \to 0} {x}^{2} / \left({x}^{2} + 0\right) = \frac{0}{0}$

since the direct compensation product equal $\frac{0}{0}$ we will use
L'Hôpital's rule

${\lim}_{\left(x\right) \rightarrow \left(a\right)} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

${\lim}_{x \to 0} {x}^{2} / \left({x}^{2} + 0\right) = {\lim}_{x \to 0} \frac{2 x}{2 x} = \frac{0}{0}$

since the direct compensation product also equal $\frac{0}{0}$ we will use
L'Hôpital's rule again

${\lim}_{x \to 0} \frac{2 x}{2 x} = {\lim}_{x \to 0} \frac{2}{2} = 1$

Now, consider approaching,(0,0) along the y-axis This means fixing x=0 and finding the limit $\textcolor{red}{{\lim}_{y \to 0} {x}^{2} / \left({x}^{2} + {y}^{2}\right)}$

${\lim}_{y \to 0 , \text{ } x = 0} {x}^{2} / \left({x}^{2} + {y}^{2}\right) = {\lim}_{y \to 0} \frac{0}{0 + {y}^{2}} = \frac{0}{0}$

${\lim}_{y \to 0} \frac{0}{0 + {y}^{2}} = {\lim}_{y \to 0} \frac{0}{2 y} = {\lim}_{y \to 0} \frac{0}{2} = 0$

Approaching the origin along these two different paths leads to different limits

$\textcolor{b l u e}{{\lim}_{x \to 0 , \text{ "y=0} x^2/(x^2+y^2)!=lim_{y->0," } x = 0} {x}^{2} / \left({x}^{2} + {y}^{2}\right)}$

Thus, the original limit does not exist.