The limit of $n^2 f \left(a-\frac{b}{n}, b+\frac{a}{n} \right)$. Let $f : \mathbb{R}^2\to \mathbb R$ be $C^2$ class function s.t. $$f(x,y)=0 \ \mathrm{ and } \ (f_x(x,y))^2+(f_y(x,y))^2=1$$ for $(x.y)\in C:=\{ (x,y) \mid x^2+y^2=1 \},$ and suppose $f$ is monotonically increasing as getting away from $(0,0)$ on the half line from $(0,0)$ to arbitrary direction.
Then, I have to calculate
$$\lim_{n\to \infty} n^2 f\left(u-\dfrac{v}{n}, v+\dfrac{u}{n}\right)$$
where $(u,v)\in C.$
At the pre-step of this problem, I had to prove
(i) $vf_x(u,v)=uf_y(u,v) $
(ii) $f_x(u,v)=u, f_y(u,v)=v $
(iii)
$\begin{pmatrix}
f_{xx}(u,v) & f_{xy}(u,v) \\
f_{yx}(u,v) & f_{yy}(u,v) \\
\end{pmatrix}
\begin{pmatrix}
-v \\
u \\
\end{pmatrix}
=\begin{pmatrix}
-v \\
u \\
\end{pmatrix}$
Thus I can use these (i)(ii)(iii) to calculate $\displaystyle\lim_{n\to \infty} n^2 f\left(u-\dfrac{v}{n}, v+\dfrac{u}{n}\right)$, but I don't know how.

Letting $\dfrac{1}{n}=m,$ I get $$\lim_{n\to \infty} n^2 f\left(u-\dfrac{v}{n}, v+\dfrac{u}{n}\right)=\lim_{m\to 0}\dfrac{f(u-vm, v+um)}{m^2}.$$
This appears to be what is relative to derivative, but I couldn't proceed.
Thanks for your help.
 A: Hint:  If you choose $f(x, y) = \frac{1}{2}x^2+\frac{1}{2}y^2-\frac{1}{2}$, then you see it satisfies all the required conditions. Observe we have that
\begin{align}
\lim_{n\rightarrow \infty} n^2f(a-\frac{b}{n}, b+\frac{a}{n})=&\ \lim_{n\rightarrow \infty} \frac{1}{2}n^2\left((a-\frac{b}{n})^2+(b+\frac{a}{n})^2 -1\right)\\
=&\ \lim_{n\rightarrow \infty} \frac{1}{2}n^2\left( a^2+b^2 + \frac{1}{n^2}(a^2+b^2)-1 \right)\\
=&\ \frac{1}{2}(a^2+b^2). 
\end{align}
In general, by Taylor's theorem, we have that
\begin{align}
f(x, y) = f(a, b) + f_x(a, b)(x-a)+f_y(a, b)(y-b) + \frac{1}{2}f_{xx}(a, b)(x-a)^2+ f_{xy}(a, b)(x-a)(y-b) + \frac{1}{2}f_{yy}(a, b)(y-b) + R_2(a, b, x, y).
\end{align}
where $|R_2| \le C((x-a)^2+(y-b)^2)^{3/2}$.
Finally, we have that
\begin{align}
f(a-\frac{b}{n}, b+\frac{a}{n}) = -\frac{b}{n}f_x(a, b)+\frac{a}{n}f_y(a, b)+ \frac{1}{2}f_{xx}(a, b)\frac{b^2}{n^2}-f_{xy}(a, b)\frac{ab}{n^2}+\frac{1}{2}f_{yy}(a, b)\frac{a^2}{n^2}+R_2(a, b, a-\frac{b}{n},b+\frac{a}{n})
\end{align}
which means
\begin{align}
n^2f(a-\frac{b}{n}, b+\frac{a}{n})=&\  -nbf_x(a, b)+naf_y(a, b)+ \frac{1}{2}f_{xx}(a, b)b^2-f_{xy}(a, b)ab+\frac{1}{2}f_{yy}(a, b)a^2+n^2R_2(a, b, a-\frac{b}{n},b+\frac{a}{n})\\
=&\ n (-b, a)\cdot (f_x(a, b), f_y(a, b))+ \frac{1}{2}f_{xx}(a, b)b^2-f_{xy}(a, b)ab+\frac{1}{2}f_{yy}(a, b)a^2+n^2R_2(a, b, a-\frac{b}{n},b+\frac{a}{n}).
\end{align}
Notice that
\begin{align}
(-b, a)\cdot (f_x(a, b), f_y(a, b)) = 0
\end{align}
(Why?).
Next, also notice that
\begin{align}
n^2|R_2|\le Cn^2\frac{1}{n^3}\rightarrow 0 
\end{align}
as $n\rightarrow \infty$.
Hence, we have that
\begin{align}
\lim_{n\rightarrow\infty} n^2f(a-\frac{b}{n}, b+\frac{a}{n}) = \frac{1}{2}[-b, a]
\begin{bmatrix}
f_{xx}(a, b) & f_{xy}(a, b)\\
f_{yx}(a, b) & f_{yy}(a, b)
\end{bmatrix}
\begin{bmatrix}
-b\\ 
a
\end{bmatrix}.
\end{align}
Combine with your (i), (ii) and (iii) yields the desired result of
\begin{align}
\lim_{n\rightarrow\infty} n^2f(a-\frac{b}{n}, b+\frac{a}{n})  = \frac{1}{2}(a^2+b^2)
\end{align}
