1

How to show this function's continuity?
My book's 'Hint' says $|xy| \leq \frac12(x^2+y^2)$ can be used.

$ f(x,y) = \left\{ \begin{array}{l l} \frac{xy}{\sqrt{x^2+y^2}} & \quad , \quad(x,y)\neq(0,0)\\ 0 & \quad , \quad(x,y)=(0,0) \end{array} \right.$

learner
  • 6,726
plhn
  • 631
  • 1
    Can you use the hint to show that $$\lim_{(x,y) \to (0,0)} \left | \frac{xy}{\sqrt{x^2+y^2}} \right | = 0$$ That is all you need to prove, since the function is continuous when $(x,y)\neq (0,0)$ quite obviously. – Prahlad Vaidyanathan Nov 18 '13 at 17:09
  • @PrahladVaidyanathan That's what I want to know about. How to show that? – plhn Nov 18 '13 at 17:11
  • Well, what happens when you apply the inequality in the hint? – Prahlad Vaidyanathan Nov 18 '13 at 17:12
  • I am not very familiar with the inequality given in the hint.. Is is $|xy|\leq \frac{1}{2}. (x^2+y^2)$ or $|xy|\leq \frac{1}{2.(x^2+y^2)}$ –  Nov 18 '13 at 17:16
  • @PraphullaKoushik : It is $|xy| \leq \frac{1}{2}(x^2+y^2)$, which follows from the fact that $(x-y)^2 \geq 0$ and $(x+y)^2 \geq 0$ – Prahlad Vaidyanathan Nov 18 '13 at 17:18
  • $\frac{xy}{\sqrt{x^2+y^2}} \leq \frac{|xy|}{\sqrt{x^2+y^2}} \leq \frac{\frac{1}{2}\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}} = \frac{1}{2}$ ... I think I should get 0 not 1/2 – plhn Nov 18 '13 at 17:20
  • 1
    ok, ok if it is, $|xy| \leq \frac{1}{2}(x^2+y^2)$ then that should give $\frac{|xy|}{\sqrt{x^2+y^2}}\leq \frac{1}{2} \sqrt{x^2+y^2}$ and the next step by considering limits would give $\lim_{(x,y) \to (0,0)} \left | \frac{xy}{\sqrt{x^2+y^2}} \right | = 0$ –  Nov 18 '13 at 17:21
  • Oh... huge mistake. Thank you two guys! very much! – plhn Nov 18 '13 at 17:23
  • When we have a relative of $x^2+y^2$ in the denominator, polar coordinates are a near-universal tool. – André Nicolas Nov 18 '13 at 17:37

3 Answers3

3

Here is another way to prove the continuity of $f(x,y)$ at $(0,0)$.

$$\left|\frac{xy}{\sqrt{x^2+y^2}}-0\right|={{|x||y|}\over{\sqrt{x^2+y^2}}} <{{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\over{\sqrt{x^2+y^2}}}=\sqrt{x^2+y^2} < \varepsilon \\\,\,(\text{where}\,\,\, \varepsilon \,\text{is a preassigned positive number})$$ if $x^2+y^2 < \delta^2,$ where $\delta =\varepsilon$. So,given any $\varepsilon >0, \exists \delta >0 $ such that $$|f(x,y)-f(0,0)|<\varepsilon \forall (x,y) \in x^2+y^2< \delta^2$$ and so $f(x,y)$ is continuous at $(0,0)$.

learner
  • 6,726
2

Even simpler ( hints can be used...but it is not compulsory!): using polar coordinates

$$\lim_{\rho\rightarrow 0}\frac{\rho^2\cos\theta\sin\theta}{\rho}= \lim_{\rho\rightarrow 0}\rho\cos\theta\sin\theta=0, $$

as we have the product of a bounded function and a function going to zero in the $\rho\rightarrow 0$ limit.

Avitus
  • 14,018
  • 1
    This argument is FALSE. You can only use polar conversion to show that the limit does NOT exist. In your case, it is still INCONCLUSIVE. There are examples where polar conversion shows the answer is zero, but approaching (0,0) according to some weird path, the limit does NOT exist. So you ought to delete your answer. – imranfat Feb 22 '21 at 01:33
  • @imranfat Can you give an example? I think , Avitus's answer is fine. As he basically showed $\lim_{\sqrt{x^2+y^2}\to0} \frac{xy}{\sqrt{x^2+y^2}}=0$ and $\sqrt{x^2+y^2}\to0\implies x\to0$ and $y\to0$. If you are talking of limits like $\lim_{r\to0} \frac {r\cos\theta} {\sin\theta}$ then I think the problem here is the limit doesn't exist when when you go to origin along the lines $\theta=2n\pi$. – PNDas Mar 02 '21 at 05:27
  • 1
    @PNDas Here is a counter example: $z=\frac{x^3}{y^2+x^3}$. With polar conversion we get $\frac{rcos^3t}{sin^2t+rcos^3t}$. With $r→0$ the answer is unequivocally $0$. So the limit is zero? No, because when we approach the Origin along the path $(t^2,t^3)$ we find for the limit $z=0.5$. Conclusion: The limit does not exit. – imranfat Mar 02 '21 at 14:42
  • @imranfat , Actually, no. If you approach the origin along, $\theta=2n\pi$, limit is 1 and there obviously exist other paths where limit is 0. So you can't say that "when $r\to0$ limit is $0$. I don't this is a valid counterexample. – PNDas Mar 02 '21 at 15:35
  • 1
    @PNDas. The point I am trying to make is that with Polars, in my example taking $r→0$ the answer is zero. (Pull the $r$ upfront the fraction). What you are doing is reasoning the other way around. First showing that through a different path the limit is NOT zero and then arguing that for $r→0$ the limit can't be zero. This is exactly what I mean: You can ONLY use the polar approach to show that the limit does NOT exist. If for $r→0$ the limit is zero, it is inconclusive at this point and thus other methods need to be explored, like for example an epsilon delta proof. – imranfat Mar 02 '21 at 16:39
0

Try to use the squeeze theorem.

Allan
  • 41