MHB What is the limit of a circumference as the power approaches infinity?

  • Thread starter Thread starter Julio1
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary
The discussion centers on evaluating the limit of the circumference as the power approaches infinity, specifically showing that the limit of C_p as p approaches infinity equals C_∞. Participants argue that as p increases, the expression for C_p simplifies to the maximum of the absolute values of x and y. There is a debate about the validity of using the ε-N definition in this context, with some skepticism about the claim that the limit is infinite. A key point raised is that if |x| is less than 1, then the limit of |x|^p approaches zero, which challenges the initial assumption. Ultimately, the conversation highlights the complexities involved in understanding the behavior of C_p at extreme values of p.
Julio1
Messages
66
Reaction score
0
Show that $\displaystyle\lim_{p\to +\infty}C_p ((0,0); 1)=C_{\infty}((0,0);1).$

Hello, I think that this limit is infinite. At $C_p=\{(x,y)\in \mathbb{R}^2: |x|^p+|y|^p=1, \, p>1\}$, then is reasonably think. But how can show this?
 
Physics news on Phys.org
An expression equivalent is $\displaystyle\lim_{p\to +\infty} \left|x \right|^p+\left|y \right|^p=\max\{\left|x \right|, \left|y \right|\}.$

Can use the definition $\varepsilon-N$?
 
Julio said:
An expression equivalent is $\displaystyle\lim_{p\to +\infty} \left|x \right|^p+\left|y \right|^p=\max\{\left|x \right|, \left|y \right|\}.$
I don't think this is true. If $|x|<1$, then $\lim_{p\to\infty}|x|^p=0$.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K