6
$\begingroup$

Find limit or show that it does not exist:

$$\lim_{(x,y) \to (0,0)} \frac{ 2x^{2}y^{3/2} }{y^{2}+x^{8}}$$ using the path $x=m y^{1/4}$:

$$\lim_{(my^{1/4},y) \to (0,0)} \frac{ 2m^{2}y^{1/2}y^{3/2}}{y^{2}+m^{8}y^{2}}$$ $$\lim_{(my^{1/4},y) \to (0,0)} \frac{ 2m^{2}y^{2}}{y^{2}(1+m^{8})}$$ $$\lim_{(my^{1/4},y) \to (0,0)} \frac{ 2m^{2}}{1+m^{8}}$$

Does not exist, because limit is path dependent.

I am asking because I have a test in a couple days. I want to make sure I'm not committing some mortal math sin. Also, the Latex tutorial is great, I think I just wasted 20 minutes making that mess.

$\endgroup$
  • 2
    $\begingroup$ Looks good to me! Good job. Intuitively you can guess that the limit won't exist because the overall power in the denominator dominates the power of the numerator. To realize this, you can look at straight-line paths to the origin by setting $y = ax$ and taking a limit that way. The numerator will have degree $\dfrac{7}{2}$ but the denominator will have degree $8$. $\endgroup$ – Cameron Williams Jun 23 '14 at 3:25
  • $\begingroup$ Thanks, I will step back and look at these problems like that before I choose an approach. I've been taking way too long to solve these. $\endgroup$ – Brett Jun 23 '14 at 3:30
2
$\begingroup$

Looks good to me! Good job. Intuitively you can guess that the limit won't exist because the overall power in the denominator dominates the power of the numerator. To realize this, you can look at straight-line paths to the origin by setting $y=ax$ and taking a limit that way. The numerator will have degree $\dfrac72$ but the denominator will have degree $8$.

Cameron Williams

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.