# Math Help - show a function of two variables is continuous

1. ## show a function of two variables is continuous

"show that $f = \frac{xy^2}{x^2+y^2}$ is continuous at (0,0), and that all directional derivatives exist at (0,0)". how would i go about showing continuity? the denominator is 0 and i guess i have to do something about that? i can't seem to cancel it down. i'm really stuck for ideas.

2. ## Re: show a function of two variables is continuous

I had this question on an old Hw. Here is the solution i gave for it. The function was

$\frac{x^2y}{x^2+y^2}$ instead of what you posted but its the same thing, switch the components in the proof.

As for directional derivatives.

Show that $lim_{h \to 0} \frac{f((0,0) + (v_1,v_2)*h) - f((0,0))}{h}$ exists.

3. ## Re: show a function of two variables is continuous

hi,
i think that answer you attached is a little too advanced for what we're doing. there's a lot of notation we haven't covered. thanks for your help, but is it possible there's a simpler way of looking at this? i totally don't follow, the x^j, E^2 etc. i follow that showing the limit using the epsilon delta definition can show continuity. perhaps what you've done is the same thing but i'm just not understanding it.

4. ## Re: show a function of two variables is continuous

i have that:

$\lim_{(x,y) \to(x_0,y_0)} f(x,y)=L \implies$ for any epsilon > 0, we can find delta > 0 such that $|f(x,y)-L|$ < epsilon. whenever $\sqrt{(x-x_0)^2+(y-y_0)^2}$ < delta. is that in any way analagous to what you've shown here?

and then show that L = 0, show that the function at (0,0) = 0, and thus that implies contunuity?

5. ## Re: show a function of two variables is continuous

$X^j = (x_1,x_2)$ where $x_1 \in \mathbb{R}, x_2 \in \mathbb{R}$ $\mathbb{E}^2 = \mathbb{R}^2$ (with the "euclideean metric" or just the normal way you measure the distance of two points that you learned in highschool algebra)
Its just a sequence from $\mathbb{R}^2$ (your domain) which converges to 0 as $j \to \infty$

6. ## Re: show a function of two variables is continuous

Maybe this is a little easier for the continuity:

First for any x,y, $|xy|\leq 1/2(x^2+y^2)$: Use the fact that $(x+y)^2\geq0$ and $(x-y)^2\geq0$

So for any $(x,y)\ne(0,0), 0\leq|{x^2y\over x^2+y^2}|\leq1/2|x|}$

By the "pinching" theorem, $\lim_{(x,y)\rightarrow(0,0)}{x^2y\over x^2+y^2}=0$

7. ## Re: show a function of two variables is continuous

i have the continuity down. can someone help with the directional derivatives? i have to show that they all exist but there are infinite... i can't show infinite things... i will run out of paper XD

8. ## Re: show a function of two variables is continuous

Have you all forgotten that a function is only continuous if the limit exists at that point and IS EQUAL TO THE FUNCTION VALUE at that point? This function is undefined at that point, so the function is clearly NOT continuous there.

However the function \displaystyle \begin{align*} f(x) = \begin{cases} \frac{x\,y^2}{x^2 + y^2} \textrm{ if } x \neq 0 \\ 0 \textrm{ if } x = 0 \end{cases} \end{align*} IS continuous at (0, 0).

9. ## Re: show a function of two variables is continuous

Prove It - my apologise for lack of clarity. f IS defined as 0 at (0,0). This is something I forgot to include, I understand the relevance.

10. ## Re: show a function of two variables is continuous

Take any non zero vector v = $(v_1,v_2)$
the directional derivative at (0,0) is defined as
$lim_{h \to 0} \frac{f( (0,0) + (v_1,v_2)h) - f(0,0)}{h} = \frac{f(v_1h,v_2h)}{h} = lim_{h \to 0} \frac{\frac{v_1h*(v_2)^2*h^2}{h^2*(v_1)^2 + h^2*(v_2)^2}}{h} = lim_{h \to 0} \frac{h^3*v_1*(v_2)^2}{h^3*((v_1)^2+(v_2)^2)} = lim_{h \to 0} \frac{v_1*(v_2)^2}{(v_1)^2 + (v_2)^2}$

whose limit exists since v is non zero and thus the directional derivative exists.

11. ## Re: show a function of two variables is continuous

so i get it, you need to show that the condition for existance is applicable to all value of v. therefore for any choice of v, the limit exists, i.e. the directional derivative exists, this is the concept?

12. ## Re: show a function of two variables is continuous

Originally Posted by learning
so i get it, you need to show that the condition for existance is applicable to all value of v. therefore for any choice of v, the limit exists, i.e. the directional derivative exists, this is the concept?
yes

13. ## Re: show a function of two variables is continuous

excellent, thanks very much. finally, one part of the question i forgot to mention: how can i show $\lim_{(x,y)\to(0,0)}\frac{xy^2}{(x^2+y^2)\sqrt{x^2 +y^2}}$ does not exist? should i look at approaching along different directions? will that happen?

to show indifferentiably, is it sufficient to show the limit in any direction is nonzero??

14. ## Re: show a function of two variables is continuous

Originally Posted by learning
excellent, thanks very much. finally, one part of the question i forgot to mention: how can i show $\lim_{(x,y)\to(0,0)}\frac{xy^2}{(x^2+y^2)\sqrt{x^2 +y^2}}$ does not exist? should i look at approaching along different directions? will that happen?

to show indifferentiably, is it sufficient to show the limit in any direction is nonzero??
Probably best to convert to polars. Remember that \displaystyle \begin{align*} x = r\cos{(\theta)}, y = r\sin{(\theta)}, x^2 + y^2 = r^2 \end{align*}, then

\displaystyle \begin{align*} \lim_{(x,y) \to (0,0)} \frac{ x\,y^2 }{ \left( x^2 + y^2 \right) \sqrt{x^2 + y^2} } &= \lim_{r \to 0} \frac{ r \cos{(\theta)} \left[ r\sin{(\theta)} \right] ^2}{r^2 \sqrt{r^2}} \\ &= \lim_{r \to 0} \frac{r^3 \cos{(\theta)} \sin^2{(\theta)} }{ r^3 } \\ &= \lim_{r \to 0} \left[ \cos{(\theta)}\sin^2{(\theta)} \right] \\ &= \cos{(\theta)}\sin^2{(\theta)} \end{align*}

Obviously this value changes depending on the value of \displaystyle \begin{align*} \theta \end{align*}, so approaching from different paths gives different values. Therefore the limit does not exist.

15. ## Re: show a function of two variables is continuous

hi,
i follow, but polar coords are not assessed on my course. i believe i am supposed to show it without any such conversion (even if it is totally valid). is there any way to do it sticking with xs and ys?

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