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Prove the limit exists for xy/(sqrt(x^2+y^2)

  • #1

Homework Statement


Find the limit, if it exists, or show that the limit does not exist.
Stewarts Calculus 7th edition 14.2 Q13.


Homework Equations





The Attempt at a Solution



[itex]f(x,y)=\frac{xy}{\sqrt{x^{2}+y^{2}}}[/itex]

The limit along any line through (0,0) is 0, as well as along other
paths through (0,0) such as [itex]x=y^{2}[/itex] and [itex]y=x^{2}[/itex]

Let [itex]\epsilon>0[/itex] We want to find [itex]\delta>0[/itex] such that

if [itex]0<\sqrt{x^{2}+y^{2}}<\delta[/itex] then [itex]\left|\frac{xy}{\sqrt{x^{2}+y^{2}}}-0\right|<\epsilon[/itex]

Edit: Still can't get the latex working
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
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Change to polar coordinates. That way the distance from (0, 0) depends only on r. If the limit, as r goes to 0, is the same, for all [itex]\theta[/itex], then the limit exists and is that value.
 
  • #3
So then [tex]f(x,y)=\frac{r\cos\theta r\sin\theta}{r^{2}}=r\cos\theta\sin\theta[/tex]

That's pretty cool! So I can just do this for most functions right?

Unfortunately, I have to prove the above using the definition of limits (as that's what we covered) for exams.
 
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  • #4
HallsofIvy
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Well, the definition says, given [itex]\epsilon> 0[/itex], there exist [itex]\delta>0[/itex] such that if the [itex]\sqrt{x^2+ y^2}= r< \delta[/itex] then [itex]|f(x, y)|< \delta[/itex]. In polar coordinates, as you show, that is [itex]|r cos(\theta)sin(\theta)|< \epsilon[/itex] and you know that [itex]cos(\theta)|\le 1[/itex] and [itex]sin(\theta)\le 1[/itex].
 

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