show-that-f-x-y-0-x-y-0-0-x-2-y-x-6-2y-2-x-y-0-0-has-a-directional-derivative-in-the-direction-of-an-arbitrary-unit-vector-at-0-0

Question Number 192138 by Mastermind last updated on 09/May/23

show that

f (x, y) = {_{0 (x, y) = (0, 0)}^{\frac{x^{2} y}{x^{6} + {2 y}^{2}} (x, y) \neq (0, 0)}

has a directional derivative in the

direction of an arbitrary unit vector

ϕ at (0, 0), but f is not continous at (0, 0)

Answered by gatocomcirrose last updated on 09/May/23

\frac{\partial f}{\partial x} (0, 0) = \lim_{x \to 0} \frac{f (x, 0) - f (0, 0)}{x - 0} = 0

\frac{\partial f}{\partial y} (0, 0) = \lim_{y \to 0} \frac{f (0, y) - f (0, 0)}{y - 0} = 0

\Rightarrow ▽ f (0, 0) = (\frac{\partial f}{\partial x} (0, 0), \frac{\partial f}{\partial y} (0, 0)) = (0, 0)

\Rightarrow directional derivative =

▽ f (0, 0) . ϕ = (0, 0) . ϕ = 0

\lim_{(x, y) \to (0, 0)} \frac{x^{2} y}{x^{6} + {2 y}^{2}} = \underset{x \to 0}{\lim f} (x, 0) = 0 = \underset{y \to 0}{\lim f} (0, x)

\overset{y = x^{2}}{=} \lim_{x \to 0} \frac{x^{4}}{x^{6} + {2 x}^{4}} = \lim_{x \to 0} \frac{1}{x^{2} + 2} = \frac{1}{2}, contradiction

\Rightarrow f is not continuous at (0, 0)

Commented by Mastermind last updated on 14/May/23

Thank you so much BOSS