How can I compute such a limit: $$\lim_{(x,y) \to(0,0)} \frac{ \ln(x+e^y)-x-y}{\sqrt{x^2+y^2}}$$?


closed as off-topic by Davide Giraudo, Daniel W. Farlow, Michael Grant, Tom-Tom, pjs36 Nov 22 '15 at 23:25

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, Daniel W. Farlow, Michael Grant, Tom-Tom, pjs36
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ If you set x to 0, the function is identically 0 for all values of y. Now, set y to 0 and take the limit as x -> 0. If it's also 0, the limit exists and is 0. If that limit isn't 0, the limit does not exist. $\endgroup$ – barrycarter Nov 22 '15 at 15:16
  • 4
    $\begingroup$ @barrycarter: That is not enough to conclude that the limit exists. For example, $$ \lim_{(x,y)\to(0,0)} \frac{xy}{x^2+y^2} $$ passes your two tests but the limit does not exist. $\endgroup$ – Henning Makholm Nov 22 '15 at 15:22
  • $\begingroup$ @HenningMakholm You're right, I'm wrong. Of course, the limit has to exist from every direction. I should've said that those two limits existing is a necessary condition, but not a sufficient one. $\endgroup$ – barrycarter Nov 22 '15 at 15:29

Hint: $\log(x+e^y)$ is differentiable at $(0,0)$ with gradient $(1,1)$, so $$ \log(x+e^y) = x+y+o(x,y) $$

  • $\begingroup$ So this limit is 0? $\endgroup$ – mrnobody Nov 22 '15 at 15:50
  • $\begingroup$ @dreamer94: Yes. In fact, in some formulations, this limit being $0$ is chosen as definition of what it means for $\log(x+e^y)$ to be differentiable with gradient $(1,1)$ -- and we then prove separately that functions of several variables are differentiable if their partial deriviatives are nice enough (which they are here). $\endgroup$ – Henning Makholm Nov 22 '15 at 15:51
  • $\begingroup$ I see, thanks for help and spelling out. $\endgroup$ – mrnobody Nov 22 '15 at 16:03

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