Htykuut

Let $f(x,y)=
begin{cases}
frac{2x^2y}{x^2+y^2} & if & (x,y) neq (0,0) \
0 & if & (x,y)=(0,0) .
end{cases}$

(i) Prove the directional derivatives of $f$ exist in any direction at the point $(0,0)$.

(ii)¿Is $f$ continuous on (0,0)?

For (i) I took $u=(u_{1},u_{2}) in mathbb{R^{2}}$ such $||u||=1$ and $0=(0,0)$. So $$lim_{t to 0} frac{f(0+ t u)-f(0)}{t}=lim_{t to 0}frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=lim_{t to 0}frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}$$. But I cannot find the limit when $t$ aproaches to $0$. Which by finding them I would prove the directional derivative exist at every direction right?

For (ii) I got the intuition $f$ is not continuous at $(0,0)$ so I took $lbrace (frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}$ and $lbrace (frac{1}{k},0) rbrace_{k in mathbb{N}}$ which are two different sequences in $mathbb{R}^{2}$ converging to $(0,0)$. However, $lbrace f(frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}=lbrace frac{1}{k^{2}} rbrace_{n in mathbb{N}} to 0$ and $lbrace f(frac{1}{k},0) rbrace_{k in mathbb{N}} to 0$. So maybe my intuition was not right? Can anyone help me end the proof of continuity or not continuity , please?

Directional derivative is defined as $$D_vec{u}f(x,y)=lim_{hto0}dfrac{f(x+ah,y+bh)-f(x,y)}{h} $$ where $vec{u}=langle a,brangle$ is the unit vector along which rate of change of $f(x,y)$ is calculated.

$$displaystyle D_vec{u}f(0,0)=lim_{hto0}dfrac{dfrac{2(ah)^2(bh)}{(ah)^2+(bh)^2}-0}{h}= lim_{hto0}frac{2a^2b}{a^2+b^2}$$ which exists since $||vec{u}||=1implies a^2+b^2neq0$. So directional derivatives of $f$ exist in any direction at the point $(0,0)$ since limit is independent of how $h$ approaches $0$.

To calculate continuity let $(x,y)to(0,0)$ along $y=mx$ where $m$ is a constant.

$$displaystylelim_{(x,mxto(0,0))}dfrac{2x^2mx}{x^2(1+m^2)}=lim_{substack{xto 0\text{along y=mx}}}dfrac{2mx}{(1+m^2)}=0=f(0,0)$$

So $f$ is continuous at $(0,0)$.