When do these limits exist?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 2K views
pawlo392
Messages
7
Reaction score
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.
 
Physics news on Phys.org
pawlo392 said:
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]?
 
  • Like
Likes   Reactions: pawlo392
Yes. Now I know. When [tex]v_1=0[/tex] this limit will equal to zero.
 
pawlo392 said:
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.
 
Mark44 said:
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.
 
pawlo392 said:
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##.
 
  • Like
Likes   Reactions: pawlo392