Partial derivative and limits

  • #1
5
0
Thread moved from the technical forums, so no Homework Template is shown
Hello . I have problems with two exercises .
1.[tex]\lim_{t \to 0 } \frac{2v_1-t^2v_2^2}{|t| \sqrt{v_1^2+v_2^2} }[/tex]
Here, I have to write when this limit will be exist.
2.[tex]\lim_{(h,k) \to (0,0) } \frac{2hk}{(|h|^a+|k|^a) \cdot \sqrt{h^2+k^2} }[/tex]
Here, I have to write for which [tex] a \in \mathbb{R}_+[/tex] this limit will equal to zero.
I don't have ideas how to do it.
 

Answers and Replies

  • #2
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,919
2,904
Hello . I have problems with two exercises .
1.[tex]\lim_{t \to 0 } \frac{2v_1-t^2v_2^2}{|t| \sqrt{v_1^2+v_2^2} }[/tex]
Here, I have to write when this limit will be exist.

Well, in a fraction, as the denominator approaches zero, then the fraction becomes undefined, unless the numerator also approaches zero. So under what circumstances does the numerator go to zero as [itex]t \rightarrow 0[/itex]?
 
  • #3
5
0
Yes. Now I know. When [tex] v_1=0[/tex] this limit will equal to zero.
 
  • #4
35,238
7,058
Yes. Now I know. When [tex] v_1=0[/tex] this limit will equal to zero.
But the limit is as t approaches 0. As far as the limit process is concerned, ##v_1## is just some constant. You can't arbitrarily say it's zero.
 
  • #5
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,919
2,904
But the limit is as t approaches 0. As far as the limit process is concerned, ##v_1## is just some constant. You can't arbitrarily say it's zero.

The question was when (in what circumstances) the limit exists. When [itex]v_1 = 0[/itex] is a possible circumstance.
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
Hello . I have problems with two exercises .
1.[tex]\lim_{t \to 0 } \frac{2v_1-t^2v_2^2}{|t| \sqrt{v_1^2+v_2^2} }[/tex]
Here, I have to write when this limit will be exist.
2.[tex]\lim_{(h,k) \to (0,0) } \frac{2hk}{(|h|^a+|k|^a) \cdot \sqrt{h^2+k^2} }[/tex]
Here, I have to write for which [tex] a \in \mathbb{R}_+[/tex] this limit will equal to zero.
I don't have ideas how to do it.

For the second one, I would use polar coordinates ##h = r \cos \theta, k = r \sin \theta##, so that we are taking the limit as ##r \to 0##.
 

Related Threads on Partial derivative and limits

  • Last Post
Replies
11
Views
3K
Replies
3
Views
6K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
10
Views
801
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
616
Top