I am very stuck on this question on finding a particular delta that would finish the proof of this limit for multi variable function. Prove that $ \displaystyle \lim_{(x,y)→(0,0)} (5x^{3}-x^{2}y^{2})=0$

I don't know how I can bound this function after I factor out $x^{2}$ from the function...

I know this is a polynomial function and all polynomial functions are continuous on $\mathbb{R}^{2}$ so we can just directly substitute stuff in but need to prove using epsilon - delta technique.

  • $\begingroup$ Using $|5x-y^2|\leq|5x| +|y^2|$ you can work out tour delta $\endgroup$ Oct 11 '15 at 3:24

If $(x,y) \in \mathbb{R}^{2}$ such that $|y| \leq |x|$, then $$ |5x^{3} - x^{2}y^{2}| \leq 5|x^{3}| + x^{2}y^{2} \leq 5|x^{3}| + 2x^{2} = x^{2}(5|x| + 2); $$ if $|x| \leq 1$, then $x^{2}(5|x|+2) \leq 7x^{2}$; taking any $\varepsilon > 0$, we have $7x^{2} < \varepsilon$ if $|x| < \varepsilon/\sqrt{7}$. We have proved this: for every $\varepsilon > 0$, if $|y| \leq |x| < \min \{1, \varepsilon/\sqrt{7} \}$, then $|5x^{3} - x^{2}y^{2}| < \varepsilon$.

  • $\begingroup$ What a mess. Knew that this function was quite nasty. Thanks for helping out. $\endgroup$ Oct 11 '15 at 3:39
  • $\begingroup$ That is beautiful, right? $\endgroup$
    – Megadeth
    Oct 11 '15 at 4:17

By definition, we are required to show that, for each $\epsilon>0$, there is some $\delta>0$ such that, for all points (x,y), if $|(x,y)-(0,0)|<\delta$, then $|5x^3-x^2y^2-0|<\epsilon$. In particular, we must be careful to avoid any dependencies between x and y, so as not to inadvertently miss important limit subsets in more pathological cases.

The $\delta$ inequality is equivalent to $\sqrt{x^2+y^2}<\delta$, so we may conveniently use polar coordinates to deduce our requirements, by defining $r=\sqrt{x^2+y^2}$, as well as $x=r\cos\theta$ and $y=r\sin\theta$.

As in most $\epsilon-\delta$ proofs, we start at the inequality we want to be true, then work backwards to find the necessary restrictions on $\delta$. Then we present the forwards implications using the found $\delta$.

First, let us rewrite the inequality in polar coordinates. We see that we require $|5r^3\cos^3(\theta)-r^4\cos^2(\theta)\sin^2(\theta)|<\epsilon$.

By the triangle inequality, we know that $|5r^3\cos^3(\theta)-r^4\cos^2(\theta)\sin^2(\theta)| \leq 5r^3|\cos^3(\theta)|+r^4\cos^2(\theta)\sin^2(\theta)$.

Since $\cos^2(\theta)\sin^2(\theta)\leq 1$, we also have $5r^3|\cos^3(\theta)|+r^4\cos^2(\theta)\sin^2(\theta)\leq 5r^3|\cos^3(\theta)|+r^4$.

Likewise, since $|\cos^3(\theta)|\leq 1$, we have $5r^3|\cos^3(\theta)|+r^4\leq 5r^3+r^4$.

If $r\geq 1$, then $5r^3+r^4\leq 5r^4+r^4=6r^4$. If $6r^4<\epsilon$, then $\frac{\epsilon}{6}>1$ and $r<\left(\frac{\epsilon}{6}\right)^\frac{1}{4}$.

If, on the other hand, $\frac{\epsilon}{6}<1$, then $r<1$ and $r^4+5r^3<r^3+5r^3$, and we have $6r^3<\epsilon$. It suffices to choose $r<\frac{\epsilon}{6}$ in this case.

Now we may begin the formal proof.

For each $\epsilon > 0$, let $\delta \leq \min\left(\frac{\epsilon}{6},\left(\frac{\epsilon}{6}\right)^\frac{1}{4}\right)$.

Then, starting with $|5r^3\cos^3(\theta)-r^4\cos^2(\theta)\sin^2(\theta)|$ and working through the inequalities as above, we come to the expression $5r^3+r^4$.

If $\epsilon\geq 6$, then $\frac{\epsilon}{6}\geq\left(\frac{\epsilon}{6}\right)^\frac{1}{4}$ and therefore $r<\left(\frac{\epsilon}{6}\right)^\frac{1}{4}$. Thus, $5r^3+r^4 < 5\left(\frac{\epsilon}{6}\right)^\frac{3}{4} + \frac{\epsilon}{6}$.

Since $\frac{\epsilon}{6}\geq 1$, we have $\left(\frac{\epsilon}{6}\right)^\frac{3}{4}\leq \frac{\epsilon}{6}$, so $5\left(\frac{\epsilon}{6}\right)^\frac{3}{4} + \frac{\epsilon}{6}\leq 5\frac{\epsilon}{6} + \frac{\epsilon}{6} = \epsilon$.

Likewise, if $\epsilon < 6$, then $r<\frac{\epsilon}{6}<1$ implies that $5r^3+r^4 < 5r^3 + r^3 = 6r^3 = \epsilon$.

As always, if you are overly concerned about using rectangular coordinates, we may simply replace $r$, $\cos\theta$ and $\sin\theta$ with the appropriate expressions.


Sometimes we can just be lucky. Like choose $\delta = \min \left\lbrace 1, \sqrt{\frac{ \varepsilon }{6}}\right\rbrace$, (From Scratch Work). This means that $0< |x|, \ |y| <\sqrt{x^2+y^2} <\delta \Rightarrow |5x^3-x^2y^2-0| \le |x|^2 (5|x|+|y|^2)<6|x|^2<6 \frac{\varepsilon}{6}=\varepsilon$


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