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

See the limit below...

Thanks

Printable View

- Jun 26th 2009, 02:38 PMJoAdams5000Limits 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 PMNonCommAlg
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 PMSampras
Wouldn't $\displaystyle

\frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \to \infty

$ if $\displaystyle \alpha > 2 $? - Jun 26th 2009, 05:04 PMNonCommAlg
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 PMSampras
if x and y were fixed...it would be right?

- Jun 26th 2009, 05:15 PMNonCommAlg
- Jun 26th 2009, 05:17 PMSampras
yes

- Jun 26th 2009, 05:34 PMNonCommAlg
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 PMSampras