Proof of existence of the limit of multi-variable function

[RunnyEGGY]

Hello people, this is my first post, so please forgive me for a potential breach of etiquette.

I'm having trouble understanding the proof that is used in the textbook I am given. Please forgive me for copying only certain parts of my textbook, since the text is in German.

so we have this function of two variables, x_1 and x_2



the point ( 0, 0 ) is obviously not defined, because that would result the denominator being equal to zero. The textbook then says that we can show that the " limit exists ".

Because of
(I)
the following is also true

(II)



with the term d( x , 0 ) converging to zero, that means f( x1 , x2 ) also converges to zero.

And so, the limit of the function f (with x going to zero) is equals to zero.

My problems understanding this proof:

(I) ist quite easy to understand, as the sum of squared numbers will always result in positive numbers.

However, I don't really do not understand the significance of (II). How can they even be sure, that the value of x1 is larger than that of the function itself. What does d( x , 0 ) even mean, and how does that relate to the value of the function.

Thank you for your answers

 

[RunnyEGGY]

For those of you who want to have a look at the original textbook page, I'm going to post it here.

 

[Deleted member 4993]

Hello people, this is my first post, so please forgive me for a potential breach of etiquette.

I'm having trouble understanding the proof that is used in the textbook I am given. Please forgive me for copying only certain parts of my textbook, since the text is in German.
so we have this function of two variables, x_1 and x_2
View attachment 18478
the point ( 0, 0 ) is obviously not defined, because that would result the denominator being equal to zero. The textbook then says that we can show that the " limit exists ".
Because of
View attachment 18479 (I)
the following is also true
View attachment 18480 (II)
with the term d( x , 0 ) converging to zero, that means f( x1 , x2 ) also converges to zero.
And so, the limit of the function f (with x going to zero) is equals to zero.
My problems understanding this proof:
(I) ist quite easy to understand, as the sum of squared numbers will always result in positive numbers.
However, I don't really do not understand the significance of (II). How can they even be sure, that the value of x1 is larger than that of the function itself. What does d( x , 0 ) even mean, and how does that relate to the value of the function.

Thank you for your answers

\(\displaystyle f(x_1,x_2) \ = \ x_1\frac{x_1^2}{x_1^2+x_2^2}\)

\(\displaystyle f(x_1,x_2) \ = \ x_1\frac{1}{1 + \frac{x_2^2}{x_1^2}}\)

\(\displaystyle \frac{1}{1 + \frac{x_2^2}{x_1^2}} \lt 1\)

Hence

\(\displaystyle |f(x_1,x_2)| \ \lt \ |x_1|\)

 

[RunnyEGGY]

Thank you very much! Just modifying the function by dividing the terms with x₂² makes finding the limit much easier!

 

[Steven G]

That really shouldn't have helped that much. A fraction (with positive numbers) is less than 1 when the denominator is larger than the numerator.

Staying away from 0: x₁² + x₂² > x₁². Therefore x₁² / (x₁² + x₂²) < 1