Limit of a function in two variables

[Misirlou]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

I tried the polar conversion, which was just messy and got me nowhere. Then I tried inputting for \epsilon-\delta formula: 0<\sqrt{x^{2}+y^{2}}<\delta and \frac{sin^{2}(a-b)}{\left|a\right|+\left|b\right|}-0<\epsilon so that sin^{2}(a-b)<(\left|a\right|+\left|b\right|)\epsilon
The tutorials I've seen use the \epsilon inequality and transform it into the \delta 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".

 

[SammyS]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

I tried the polar conversion, which was just messy and got me nowhere. Then I tried inputting for \epsilon-\delta formula: 0<\sqrt{x^{2}+y^{2}}<\delta and \frac{sin^{2}(a-b)}{\left|a\right|+\left|b\right|}-0<\epsilon so that sin^{2}(a-b)<(\left|a\right|+\left|b\right|)\epsilon
The tutorials I've seen use the \epsilon inequality and transform it into the \delta 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 \left|\sin(a+b)\right|\leq\left|a+b\right|\leq \left|a\right|+\left|b\right|,

then \left|\sin(a-b)\right|

=\left|\sin(a+(-b))\right|\leq\left|a+(-b)\right|\leq\left|a\right|+\left|-b\right|

               =\left|a\right|+\left|b\right|​

 

[Misirlou]

Okay, that makes more sense. So if sin^{2}(a-b)\leq sin(a-b)\leq\left|a\right|+\left|b\right|
Then
\left|a\right|+\left|b\right|<\left|a\right|+\left|b\right|\epsilon
So
1\leq \epsilon

However, I've been taught to solve for \delta first, and then prove with Given \epsilon>0 and Choose \delta= constant*\epsilon. Now if I have \epsilon first, how do I construct a proof this way ?

 

[Misirlou]

Any guidance is appreciated :)

 

[SammyS]

Okay, that makes more sense. So if sin^{2}(a-b)\leq sin(a-b)\leq\left|a\right|+\left|b\right|
Then
\left|a\right|+\left|b\right|<\left|a\right|+\left|b\right|\epsilon
So
1\leq \epsilon

However, I've been taught to solve for \delta first, and then prove with Given \epsilon>0 and Choose \delta= constant*\epsilon. Now if I have \epsilon first, how do I construct a proof this way ?

That basically shows that \displaystyle \frac{\sin^{2}(a-b)}{\left|a\right|+\left|b\right|}\leq1 no matter what the value of δ .

But you need to find a δ so that \displaystyle \frac{\sin^{2}(a-b)}{\left|a\right|+\left|b\right|}\leq\varepsilon, whenever \displaystyle 0<\sqrt{a^2+b^2}<\delta\,.

 

[Misirlou]

I'm not sure how I can go about that. Where should I start looking for a delta ?

 

[SammyS]

I'm not sure how I can go about that. Where should I start looking for a delta ?

Well, you know that \displaystyle \left|\sin(a-b)\right|\leq\left|a \right|+\left|b\right|\,,

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

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.

 

[Misirlou]

Okay, so this is where I am. 0\leq\sqrt{a^{2}+b^{2}}^{2}\leq\delta^{2}
So
0\leq\left|a^{2}+b^{2}\right|\leq\delta^{2}
and because
\left|a^{2}+b^{2}\right|\leq \left|a^{2}\right|+\left|b^{2}\right|
Then
0\leq\left|a^{2}\right|+\left|b^{2}\right|\leq\delta^{2}
and I would somehow slide in that
\frac{\sin^{2}(a-b)}{\left|a\right|+\left|b\right|}≤\left|a\right|+\left|b\right|
I feel like I could use the triangle inequality: \left|a\right|+\left|b\right|\geq \sqrt{a^{2}+b^{2}}
but I am I allowed to squeeze it in between the \sqrt{a^{2}+b^{2}}and δ?
because I've so far only squeezed stuff in between \sqrt{a^{2}+b^{2}} and 0

 

[SammyS]

Okay, so this is where I am. 0\leq\sqrt{a^{2}+b^{2}}^{2}\leq\delta^{2}
So
0\leq\left|a^{2}+b^{2}\right|\leq\delta^{2}
and because
\left|a^{2}+b^{2}\right|\leq \left|a^{2}\right|+\left|b^{2}\right|
Then
0\leq\left|a^{2}\right|+\left|b^{2}\right|\leq \delta^{2}
and I would somehow slide in that
\frac{\sin^{2}(a-b)}{\left|a\right|+\left|b\right|}≤\left|a\right|+\left|b\right|
I feel like I could use the triangle inequality: \left|a\right|+\left|b\right|\geq \sqrt{a^{2}+b^{2}}
but I am I allowed to squeeze it in between the \sqrt{a^{2}+b^{2}}and δ?
because I've so far only squeezed stuff in between \sqrt{a^{2}+b^{2}} and 0

Can you show that \displaystyle 2\sqrt{a^2+b^2}\ge\left|a\right|+\left|b\right|\ ?