The only given condition is that $f$ is a $C^1$ function defined on $\mathbb{R}^2$.
How can I find the value of $$\lim_{r \rightarrow 0}\frac{1}{\pi}\int_0 ^{2\pi}\frac{\cos \theta}{r} f(r\cos \theta, r\sin \theta)d\theta$$
I tried to interchange the coordinate into standard coordinate, but I think it is useless as so complicated.
Thank you if you give me any hints.
By the formula for directional derivatives we have $$ \lim_{r \to 0} \frac{f(r\cos\theta, r\sin\theta)-f(0,0)}{r} = (\nabla f(0,0))\cdot (\cos\theta, \sin\theta) = f_x(0,0)\cos\theta + f_y(0,0)\sin\theta. $$ Thus $$ \frac{1}{2\pi} \int_0^{2\pi} \cos\theta \Big( \frac{f(r\cos\theta, r\sin\theta)-f(0,0)}{r} + \frac{f(0,0)}{r}\Big) d \theta $$ $$ =\frac{1}{2\pi} \int_0^{2\pi} \cos\theta \Big( \frac{f(r\cos\theta, r\sin\theta)-f(0,0)}{r} \Big) d \theta $$ $$ \to_{r \to 0} \frac{1}{2\pi} \int_0^{2\pi}\cos\theta(f_x(0,0)\cos\theta + f_y(0,0)\sin\theta)d \theta $$ $$ =\frac{1}{2\pi} \int_0^{2\pi} f_x(0,0)(\cos\theta)^2 d \theta = \frac{f_x(0,0)}{2}. $$