Prove that $\lim_{t\to 0}f(g(t))$ exists, where $g$ is differentiable and $f(x,y)=x^2y/(x^2+y^2)$. Let $f:\mathbb{R}^2\to\mathbb{R}$ be a function defined by 
$$f(x,y)=\left\{\begin{matrix}
\frac{x^2y}{x^2+y^2}&\text{if }(x,y)\neq(0,0) \\ 
 0&\text{if }x=y=0 
\end{matrix}\right.$$
Let $\varepsilon>0$ and $g:(-\varepsilon,\varepsilon)\to\mathbb{R}^2$ be a function differentiable at point $0$ such that $g(0)=(0,0)$. Prove that the following limit exists. 
$$\lim_{t\to 0}\frac{f\left(g(t)\right)}{t}$$
Thanks.
 A: Let $g(t)=(g_{1}(t),g_{2}(t))$. Suppose that at least one of $g_{i}'(0)\neq0$ for $i=1,2$ then:
$\lim_{t\to0}\frac{f(g(t))}{t}=\lim_{t\to0}\frac{f(g_{1}(t),g_{2}(t))}{t}=\lim_{t\to0}\frac{\frac{g_{1}^{2}(t)g_{2}(t)}{g_{1}^{2}(t)+g_{2}^{2}(t)}}{t}=\lim_{t\to0}\frac{\big(\frac{g_{1}(t)}{t}\big)^{2}\frac{g_{2}(t)}{t}}{\big(\frac{g_{1}(t)}{t}\big)^{2}+\big(\frac{g_{2}(t)}{t}\big)^{2}}=\frac{(g'_{1}(0))^{2}g'_{2}(0)}{(g'_{1}(0))^{2}+(g'_{2}(0))^{2}}$
If both $g'_{1}(0)=0=g'_{2}(0)$ then by the arithmetic-geometric inequality:
$|g_{1}(t)g_{2}(t)|=\sqrt{|g^{2}_{1}(t)||g^{2}_{2}(t)|}\le\frac{|g_{1}(t)|^{2}+|g_{2}(t)|^{2}}{2}=\frac{|g_{1}^{2}(t)+g_{2}^{2}(t)|}{2}$.
So for points where not both $g_{i}(t)=0$ for $i=1,2$ then we have $|\frac{g_{1}(t)g_{2}(t)}{g_{1}^{2}(t)+g_{2}^{2}(t)}|\le\frac{1}{2}$.
So $|\frac{\frac{g_{1}^{2}(t)g_{2}(t)}{g_{1}^{2}(t)+g_{2}^{2}(t)}}{t}|\le\frac{|\frac{g_{1}(t)}{t}|}{2}$
which goes to $0$ since $g'_{1}(0)=0$ by assumption. If arbitrarily close there are points such that $g_{i}(t)=0$ for both $i=1,2$ then the limit along these points is also $0$ (since in this case $f(g_{1}(t),g_{2}(t))=f(0,0)=0$ by the definition of $f$). So the limit exists and is $0$ in this case.
