0
$\begingroup$

I've tried solving the following question using the squeeze theorem, changing to polar coordinates, and trying to plug in different values, but I don't seem to be able to prove that the limit exists (or doesn't). How would you go about solving this? $$ \lim_{(x,y) \to (0, 0)}\frac{1-\cos(\sqrt{xy})}{xy} $$

$\endgroup$
1
  • $\begingroup$ Expand in Taylor series $\endgroup$
    – FShrike
    Dec 12, 2022 at 18:02

1 Answer 1

Reset to default
1
$\begingroup$

You can solve this using the limit of the composition of functions... In other words, if you set $w = \sqrt{xy}$, your limit is simply $$ \lim_{w\to 0^+} \dfrac{1- \cos w}{w^2} = \lim_{w\to 0^+} \frac{\sin w}{2w} = \frac 12. $$

You must keep in mind that, in the original variables, the limit is taken along the set $D=\{(x,y)\in \mathbb{R}^2: x y \ge 0\}$.

$\endgroup$
1
  • $\begingroup$ Thank you! After a long time of not seeing any limits, jumping back into it with two variables has been qutie confusing. $\endgroup$
    – rattulus
    Dec 12, 2022 at 18:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .