I'm supposed to show whether the following limit exists or doesn't exist and if it exist calculate its value.

$$\lim_{(x,y,z) \to (0,0,0)} \frac{\ln(1+x^2+y^2+z^2)}{\sin(x^2+y^2+z^2)+xyz}$$

My approach has been to go along paths:

Along the path $x=y=z$ I get the limit to go to $1$ as follows :

$$\lim_{x \to 0} \frac{\ln(1+3x^2)}{\sin(3x^2)+x^3}$$

Using Taylor's approximation for $\ln$ and $\sin$, I simplified it to :

$$\lim_{x \to 0} \frac{3x^2}{3x^2+x^3}= 1 $$

Showing what the limit is along $1$ path is not enough to show that the limit exists. How should I continue (or start over with) this problem to officially show that the limit exists and goes to 1 or that it in fact does not?

  • $\begingroup$ Use spherical coordinates then proceed with the Taylor series expansion as you have. $\endgroup$ – Peter Foreman Jul 9 at 16:16
  • $\begingroup$ I am only familiar with polar when there is no $z$ involved. Could you please elaborate a little bit on this approach? $\endgroup$ – marky Jul 9 at 16:18
  • $\begingroup$ Using spherical coordinates (3D polar coordinates) we have that $x=r\sin{\theta}\cos{\phi}$, $y=r\sin{\theta}\sin{\phi}$ and $z=r\cos{\theta}$ where we have $r\ge0$, $\theta\in[0,\pi]$ and $\phi\in[0,2\pi)$. $\endgroup$ – Peter Foreman Jul 9 at 16:22
  • 1
    $\begingroup$ See here $\endgroup$ – DMcMor Jul 9 at 16:23
  • $\begingroup$ Ok that makes more sense. I believe, Spherical coordinates are something different from polar in 3D, polar in 3D simply means that $z$ is not redefined. What I remember from polar in 2D is that we then let $lim_{r\to 0}$. Do we do the same for spherical? $\endgroup$ – marky Jul 9 at 16:33

Your Answer

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

Browse other questions tagged or ask your own question.