Proving the Limit of f(x,y)= y+x^2cosy is 0 at (0,0) by Epsilon-Delta Definition

[sigh1342]

Homework Statement

show that $$f(x,y) =y+x^2cosy $$ has a limit 0 at (0,0) by the ε-δ definition.

Homework Equations

The Attempt at a Solution

$$|y+x^2cosy| ≤ |y|+|x^2|$$ (by tri. inequ. and $$|cosy|≤1$$
then can I suppose $$|x^2|<|x|$$ , since $$|x|<1$$,
then $$|y+x^2cosy| ≤ |x|+|y| ≤ 2\sqrt{x^2+y^2} $$ ?
if not , how can I do it by the ε-δ definition

 

[hedipaldi]

Attached

 

[HallsofIvy]

Given $\epsilon> 0$, show that there exist $\delta> 0$ such that if $\sqrt{x^2+ y^2}< \delta$ (this is the distance from (x, y) to (0, 0)) then $|y+ x^2cos(y)|<\epsilon$.

From your last inequality, $|y+ x^2cos(y)|< 2\sqrt{x^2+ y^2}$ you just need to choose $\delta< \epsilon/2$.

 

[sigh1342]

thanks you guys, actually I would want to know can I suppose x^2|<|x| , since |x|<1

 

[STEMucator]

Hmm depends on how small you want your neighborhood to be in terms of your choice of δ.

Otherwise you can assume that |x| ≤ (x²+y²)^1/2 < δ and |y| ≤ (x²+y²)^1/2 < δ

So that : $|y| + |x|^2 < δ + δ^2 = δ(δ+1)$ Then for δ≤1, we would have :

$|y| + |x|^2 < δ + δ^2 = δ(δ+1) ≤ 2δ ≤ ε$

So as long as δ ≤ min{1, ε/2} you're good.