Limit of a function in two variables

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Homework Statement



Prove with [itex]\epsilon-\delta[/itex]: [itex]Lim_{(a,b)\rightarrow(0,0)}\frac{sin^{2}(a-b)}{\left|a\right|+\left|b\right|}=0[/itex]
Hint: [itex]\left|sin(a+b)\right|\leq\left|a+b\right|\leq\left|a\right|+\left|b\right|[/itex]

Homework Equations



[itex]0<\sqrt{(x-x_{0})^{2}+(y-y_{0})^{2}}<\delta[/itex]
and
[itex]\left|f(x,y)-L\right|<\epsilon[/itex]

The Attempt at a Solution



I tried the polar conversion, which was just messy and got me nowhere. Then I tried inputting for [itex]\epsilon-\delta[/itex] formula: [itex]0<\sqrt{x^{2}+y^{2}}<\delta[/itex] and [itex]\frac{sin^{2}(a-b)}{\left|a\right|+\left|b\right|}-0<\epsilon[/itex] so that [itex]sin^{2}(a-b)<(\left|a\right|+\left|b\right|)\epsilon[/itex]
The tutorials I've seen use the [itex]\epsilon[/itex] inequality and transform it into the [itex]\delta[/itex] inequality, but I don't know how to do this. I'm just looking for a way to solve it, not necessarily using the "hint".
 
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Answers and Replies

  • #2
SammyS
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Homework Statement



Prove with [itex]\epsilon-\delta[/itex]: [itex]Lim_{(a,b)\rightarrow(0,0)}\frac{sin^{2}(a-b)}{\left|a\right|+\left|b\right|}[/itex]
Hint: [itex]\left|sin(a+b)\right|\leq\left|a+b\right|\leq\left|a\right|+\left|b\right|[/itex]

Homework Equations



[itex]0<\sqrt{(x-x_{0})^{2}+(y-y_{0})^{2}}<\delta[/itex]
and
[itex]\left|f(x,y)-L\right|<\epsilon[/itex]

The Attempt at a Solution



I tried the polar conversion, which was just messy and got me nowhere. Then I tried inputting for [itex]\epsilon-\delta[/itex] formula: [itex]0<\sqrt{x^{2}+y^{2}}<\delta[/itex] and [itex]\frac{sin^{2}(a-b)}{\left|a\right|+\left|b\right|}-0<\epsilon[/itex] so that [itex]sin^{2}(a-b)<(\left|a\right|+\left|b\right|)\epsilon[/itex]
The tutorials I've seen use the [itex]\epsilon[/itex] inequality and transform it into the [itex]\delta[/itex] inequality, but I don't know how to do this. I'm just looking for a way to solve it, not necessarily using the "hint".
The hint appears to be very useful here.

If [itex]\left|\sin(a+b)\right|\leq\left|a+b\right|\leq \left|a\right|+\left|b\right|[/itex],

then [itex]\left|\sin(a-b)\right|[/itex]
[itex]=\left|\sin(a+(-b))\right|\leq\left|a+(-b)\right|\leq\left|a\right|+\left|-b\right|[/itex]
               [itex]=\left|a\right|+\left|b\right|[/itex]​
 
  • #3
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Okay, that makes more sense. So if [itex]sin^{2}(a-b)\leq sin(a-b)\leq\left|a\right|+\left|b\right|[/itex]
Then
[itex]\left|a\right|+\left|b\right|<\left|a\right|+\left|b\right|\epsilon[/itex]
So
[itex]1\leq \epsilon[/itex]

However, I've been taught to solve for [itex]\delta[/itex] first, and then prove with Given [itex]\epsilon>0[/itex] and Choose [itex]\delta= constant*\epsilon[/itex]. Now if I have [itex]\epsilon[/itex] first, how do I construct a proof this way ?
 
  • #4
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Any guidance is appreciated :)
 
  • #5
SammyS
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Okay, that makes more sense. So if [itex]sin^{2}(a-b)\leq sin(a-b)\leq\left|a\right|+\left|b\right|[/itex]
Then
[itex]\left|a\right|+\left|b\right|<\left|a\right|+\left|b\right|\epsilon[/itex]
So
[itex]1\leq \epsilon[/itex]

However, I've been taught to solve for [itex]\delta[/itex] first, and then prove with Given [itex]\epsilon>0[/itex] and Choose [itex]\delta= constant*\epsilon[/itex]. Now if I have [itex]\epsilon[/itex] first, how do I construct a proof this way ?
That basically shows that [itex]\displaystyle \frac{\sin^{2}(a-b)}{\left|a\right|+\left|b\right|}\leq1[/itex] no matter what the value of δ .

But you need to find a δ so that [itex]\displaystyle \frac{\sin^{2}(a-b)}{\left|a\right|+\left|b\right|}\leq\varepsilon[/itex], whenever [itex]\displaystyle 0<\sqrt{a^2+b^2}<\delta\,.[/itex]
 
  • #6
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I'm not sure how I can go about that. Where should I start looking for a delta ?
 
  • #7
SammyS
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I'm not sure how I can go about that. Where should I start looking for a delta ?
Well, you know that [itex]\displaystyle \left|\sin(a-b)\right|\leq\left|a \right|+\left|b\right|\,,[/itex]

so [itex]\displaystyle \frac{\sin^{2}(a-b)}{\left| a\right|+\left|b\right|}\leq\left|a\right|+\left|b\right|\ .[/itex]

And as is usually done, start with ε and \work your way to δ.

Then for the proof you write-up, reverse the order.

Typo fixed in Edit.
 
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  • #8
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Okay, so this is where I am. [itex]0\leq\sqrt{a^{2}+b^{2}}^{2}\leq\delta^{2}[/itex]
So
[itex]0\leq\left|a^{2}+b^{2}\right|\leq\delta^{2}[/itex]
and because
[itex]\left|a^{2}+b^{2}\right|\leq \left|a^{2}\right|+\left|b^{2}\right|[/itex]
Then
[itex]0\leq\left|a^{2}\right|+\left|b^{2}\right|\leq\delta^{2}[/itex]
and I would somehow slide in that
[itex]\frac{\sin^{2}(a-b)}{\left|a\right|+\left|b\right|}[/itex]≤[itex]\left|a\right|+\left|b\right|[/itex]
I feel like I could use the triangle inequality: [itex]\left|a\right|+\left|b\right|\geq \sqrt{a^{2}+b^{2}}[/itex]
but I am I allowed to squeeze it in between the [itex]\sqrt{a^{2}+b^{2}}[/itex]and δ?
because I've so far only squeezed stuff in between [itex]\sqrt{a^{2}+b^{2}}[/itex] and 0
 
  • #9
SammyS
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Okay, so this is where I am. [itex]0\leq\sqrt{a^{2}+b^{2}}^{2}\leq\delta^{2}[/itex]
So
[itex]0\leq\left|a^{2}+b^{2}\right|\leq\delta^{2}[/itex]
and because
[itex]\left|a^{2}+b^{2}\right|\leq \left|a^{2}\right|+\left|b^{2}\right|[/itex]
Then
[itex]0\leq\left|a^{2}\right|+\left|b^{2}\right|\leq \delta^{2}[/itex]
and I would somehow slide in that
[itex]\frac{\sin^{2}(a-b)}{\left|a\right|+\left|b\right|}[/itex]≤[itex]\left|a\right|+\left|b\right|[/itex]
I feel like I could use the triangle inequality: [itex]\left|a\right|+\left|b\right|\geq \sqrt{a^{2}+b^{2}}[/itex]
but I am I allowed to squeeze it in between the [itex]\sqrt{a^{2}+b^{2}}[/itex]and δ?
because I've so far only squeezed stuff in between [itex]\sqrt{a^{2}+b^{2}}[/itex] and 0
Can you show that [itex]\displaystyle 2\sqrt{a^2+b^2}\ge\left|a\right|+\left|b\right|\ ?[/itex]
 

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