0
$\begingroup$

CONTEXT: University first year multivariate calculus maths course.

I need to find the limit of $f:ℝ^2$\ $(0,0) \to ℝ$, or show that it does not exist. $$f(x,y)=\frac{y^2x}{y^4+2x^2}$$

I know that it does not exist, but don't know how to show this (without writing a rigorous proof as that is beyond the scope of the course) by finding two different limits along two different paths within the domain.

I have checked along these paths: $x$-axis, $y$-axis, $y=x$ and so far they all give limits of $0$. I also looked along the path $x=y^2$ which gave a limit of $\frac{1}{3}$, but was unsure of whether you can consider nonlinear paths (can I use this?).

Any guidance would be greatly appreciated!

$\endgroup$
  • 1
    $\begingroup$ You're already done, any path is okay as long as it gets to the point (for example $y=x+1$ would not be acceptable) $\endgroup$ – Ninad Munshi Sep 16 '19 at 4:11
  • $\begingroup$ @NinadMunshi Sorry, what do you mean by as long it gets to the point? Does that mean that for any path to be suitable, if you're considering for example the limit as it approaches $(0,0)$, if you let $x=0$, then $y=0$ and vice versa? $\endgroup$ – Ruby Pa Sep 16 '19 at 4:20
  • 1
    $\begingroup$ Your description of the problem was incomplete. You obviously meant the limit was going to $(0,0)$. But say instead you want to approach the point $(1,1)$. You can't use the paths $x=0$ or $y=0$ because they never get to that point. $\endgroup$ – Ninad Munshi Sep 16 '19 at 4:25
1
$\begingroup$

You are right.

We have $f(x,0)=0$ for $x \ne 0$, hence $f(x,0) \to 0$ as $x \to 0.$

Furthermore $f(y^2,y)=1/3$ for $y \ne 0$, hence $f(y^2,y) \to 1/3$ as $y \to 0.$

Conclusion: $\lim_{(x,y) \to (0,0)}f(x,y)$ does not exist.

| cite | improve this answer | |
$\endgroup$

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.