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

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

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

no! put $x=r\cos \theta, \ y=r \sin \theta.$ then, since $(x,y) \to (0,0),$ we have $r=\sqrt{x^2+y^2} \to 0.$ thus: $0 \leq \frac{|x|^{\alpha}+|y|^{\alpha}}{x^2+y^2} =r^{\alpha - 2}(|\cos \theta|^{\alpha} + |\sin \theta|^{\alpha}) \leq 2 r^{\alpha -2} \to 0.$
• 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 $x,y$ were fixed and $\alpha \to \infty,$ then $\frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \to \infty ?$ 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 $x,y$: 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 $2 \cos^2 \left(\frac{\pi}{4} - \frac{|x|^{\alpha} + |y|^{\alpha}}{2(x^2+y^2)} \right)=1 + \sin \left(\frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \right).$ now consider three cases: $\alpha < 2, \ \alpha = 2, \ \alpha > 2.$ show that if $\alpha < 2,$ the limit doesn't exist. if $\alpha=2,$ then the limit is

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

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