1. ## Limits of Curves

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

See the limit below...

Thanks

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 $\displaystyle e.$

3. Wouldn't $\displaystyle \frac{|x|^{\alpha} + |y|^{\alpha}}{x^2+y^2} \to \infty$ if $\displaystyle \alpha > 2$?

4. 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: $\displaystyle 0 \leq \frac{|x|^{\alpha}+|y|^{\alpha}}{x^2+y^2} =r^{\alpha - 2}(|\cos \theta|^{\alpha} + |\sin \theta|^{\alpha}) \leq 2 r^{\alpha -2} \to 0.$

5. if x and y were fixed...it would be right?

6. 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?

7. yes

8. 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.

9. 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 $\displaystyle \alpha < 2$?