Consider $f(x,y) = x$ We must have: [ D = \left( \frac{\|f(\vec{u}) - f(\vec{v}) - L(\vec{u}-\vec{v})\|}{\|\vec{u}-\vec{v}\|}\right)^2 = \frac{\|(u1 - v1, 0) - (L1, L2)\cdot (u1 - v1, u2 - v2)\|^2}{\|\vec{u}-\vec{v}\|^2} \to 0 ] [ D = \frac{(u1-v1 - L1(u1-v1))^2 + (L2(u2-v2))^2}{(u1-v1)^2 + (u2-v2)^2} = \frac{(u1-v1 - L1(u1-v1))^2}{(u1-v1)^2 + (u2-v2)^2} + \frac{(L2(u2-v2))^2}{(u1-v1)^2 + (u2-v2)^2} ] Both terms are nonnegative, so they must approach 0 individually.

[ \frac{(u1-v1 - L1(u1-v1))^2}{(u1-v1)^2 + (u2-v2)^2} \to 0 \implies L1 = 1 ] [ \frac{(L2(u2-v2))^2}{(u1-v1)^2 + (u2-v2)^2} \to 0 \implies L2 = 0 ]

Thus $L = (1,0)$.

Friday, 17 August 2018 01:01 GMT