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 $\displaystyle 2\sqrt{a^2+b^2}\ge\left|a\right|+\left|b\right|\ ?$