# Limits of Curves

• Jun 26th 2009, 02:38 PM
Limits of Curves
Find all values of 'alpha' for which the limit exists. For these values, evaluate the limit.

See the limit below...

Thanks
• Jun 26th 2009, 04:44 PM
NonCommAlg
Quote:

Find all values of 'alpha' for which the limit exists. For these values, evaluate the limit.

See the limit below...

Thanks

use the identity $\displaystyle 2 \cos^2 \left(\frac{\pi}{4} - \frac{|x|^{\alpha} + |y|^{\alpha}}{2(x^2+y^2)} \right)=1 + \sin \left(\frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \right).$ now consider three cases: $\displaystyle \alpha < 2, \ \alpha = 2, \ \alpha > 2.$ show that if $\displaystyle \alpha < 2,$ the limit doesn't exist. if $\displaystyle \alpha=2,$ then the limit is

obviously $\displaystyle 1+\sin 1.$ finally if $\displaystyle \alpha > 2,$ then $\displaystyle \frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \to 0$ and thus in this case the limit is equal to $\displaystyle e.$
• Jun 26th 2009, 04:57 PM
Sampras
Wouldn't $\displaystyle \frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \to \infty$ if $\displaystyle \alpha > 2$?
• Jun 26th 2009, 05:04 PM
NonCommAlg
Quote:

Originally Posted by Sampras
Wouldn't $\displaystyle \frac{|x|^{\alpha}+|y|^{\alpha}}{x^2+y^2} \to \infty$ if $\displaystyle \alpha > 2$?

no! put $\displaystyle x=r\cos \theta, \ y=r \sin \theta.$ then, since $\displaystyle (x,y) \to (0,0),$ we have $\displaystyle r=\sqrt{x^2+y^2} \to 0.$ thus: $\displaystyle 0 \leq \frac{|x|^{\alpha}+|y|^{\alpha}}{x^2+y^2} =r^{\alpha - 2}(|\cos \theta|^{\alpha} + |\sin \theta|^{\alpha}) \leq 2 r^{\alpha -2} \to 0.$
• Jun 26th 2009, 05:10 PM
Sampras
if x and y were fixed...it would be right?
• Jun 26th 2009, 05:15 PM
NonCommAlg
Quote:

Originally Posted by Sampras
if x and y were fixed...it would be right?

you mean if $\displaystyle x,y$ were fixed and $\displaystyle \alpha \to \infty,$ then $\displaystyle \frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \to \infty ?$ is that what you're asking?
• Jun 26th 2009, 05:17 PM
Sampras
yes
• Jun 26th 2009, 05:34 PM
NonCommAlg
Quote:

Originally Posted by Sampras
yes

the limit will depend on $\displaystyle x,y$: if 0 < |x|, |y| < 1, the limit is 0. if |x| = |y| = 1, the limit is 1. if |x| > 1 or |y| > 1, then the limit is infinity. if x = y = 0, the limit obviously doesn't exist.

to avoid any possible confusion: Sampras' question is kind of off-topic and not related to JoAdams5000's question.
• Jun 26th 2009, 07:50 PM
Sampras
Quote:

Originally Posted by NonCommAlg
use the identity $\displaystyle 2 \cos^2 \left(\frac{\pi}{4} - \frac{|x|^{\alpha} + |y|^{\alpha}}{2(x^2+y^2)} \right)=1 + \sin \left(\frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \right).$ now consider three cases: $\displaystyle \alpha < 2, \ \alpha = 2, \ \alpha > 2.$ show that if $\displaystyle \alpha < 2,$ the limit doesn't exist. if $\displaystyle \alpha=2,$ then the limit is

obviously $\displaystyle 1+\sin 1.$ finally if $\displaystyle \alpha > 2,$ then $\displaystyle \frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \to 0$ and thus in this case the limit is equal to $\displaystyle e.$

How did you come up with the identity: $\displaystyle 2 \cos^{2} \left(\frac{\pi}{4}- \frac{\theta}{2} \right) = 1 + \sin \theta$? And how did you get $\displaystyle e$ for $\displaystyle \alpha < 2$?