- #1

- 804

- 37

## Homework Statement

"Let ##f:ℝ^2\rightarrow ℝ## be defined by ##f(x,y)=\sqrt{|xy|}##. Show that ##f## is not differentiable at ##(0,0)##."

## Homework Equations

__Differentiability:__If ##f:ℝ^n\rightarrow ℝ^m## is differentiable at ##a\in ℝ^n##, then there exists a unique linear transformation such that ##\lim_{h\rightarrow 0} \frac{f(a+h)-f(a)-\lambda (h)}{|h|} = 0##.

## The Attempt at a Solution

So far, with ##h=(b,k)##, I have the equality:

##\lim_{h\rightarrow 0} \frac{f(0+h)-f(0)-\lambda (h)}{|h|} = \lim_{h\rightarrow 0} \frac{f(h)-\lambda (h)}{|h|}=\lim_{h\rightarrow 0} \frac{\sqrt{|bk|}-\lambda (h)}{\sqrt{b^2+k^2}}=\lim_{h\rightarrow 0} \frac{\sqrt{|bk|}-\lambda (h)}{\sqrt{b^2+k^2}}=\lim_{h\rightarrow 0} \frac{\sqrt{|\frac{b}{k}|}-\frac{\lambda (h)}{k^2}}{\sqrt{(\frac{b}{k})^2+1}}##

I don't know how to prove that this limit is not zero, though. All I'm thinking about how to solve this by arguing that the only way the limit is zero, is if ##\lambda (h) = \sqrt{|bk|}##, which shouldn't be linear, since neither vector addition nor scalar multiplication with negative numbers holds.

Last edited: