Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to prove that $f$ continuous at $(x, y)=(0,0)$ using a $\epsilon$-$\delta$ proof $$ f(x, y) = \begin{cases} \frac{x^3}{{x^2 + y^2}},&(x,y)\neq (0,0) \\ 0,&(x,y) = (0,0) \end{cases} $$

I'm not sure how to manipulate the function to get $\delta$

share|cite|improve this question
up vote 2 down vote accepted

Hint: Use that $\displaystyle\frac{x^2}{x^2+y^2}\le 1$. Then start the proof like

Assume we are given an $\varepsilon>0$. Let $\delta:=$ ...

If the $(x,y)$ point is closer to $(0,0)$ than $\delta$, then, in particular, $|x|<\delta$, so ...

share|cite|improve this answer
    
the numerator is $x^3$ – PooperScooper Feb 21 '13 at 1:20
2  
Exactly. That is, $x\cdot x^2$. – Berci Feb 21 '13 at 1:21
    
oh ok ok my bad thanks! – PooperScooper Feb 21 '13 at 1:24
    
$\delta:=\varepsilon$ – Berci Feb 21 '13 at 1:35

You can use polar coordinates to see the problem more clearly: $$f(r,\theta)=\frac{r^3 \cos^3(\theta)}{r^2} = r \cos^3(\theta)$$ Thus, for any $\epsilon>0$, choose $\delta=\epsilon$. If $r<\delta$: $$|f(r,\theta)-0| < |r \cos^3(\theta)-0|<|r|<\delta=\epsilon$$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.