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

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

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

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  • $\begingroup$ 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

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