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I made the following: $$\lim_{(x,y)\rightarrow(0,0)} \frac{xy f(x,y) - 0 -g_{x}(0,0) - g_{y}(0,0) (y-0)}{((x-0)^2 + (y-0)^2)^{\frac{1}{2}}}$$ $$=\lim_{(x,y)\rightarrow(0,0)} \frac{xy f(x,y)}{(x^2+y^2)^{\frac{1}{2}}}$$

Now, my problem is how to continue to finish the demonstration.

Isaac Browne
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Ayesca
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2 Answers2

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You need to find $\lim_{(x,y)\to(0,0)}\frac{xyf(x,y)}{\sqrt{x^2+y^2}}$. Now the easiest way to find the limit in my opinion is change to polar coordinates. So $x=r\cos\theta,y=r\sin\theta$ and now the limit is:

$\lim_{r\to 0}\frac{r^2cos(\theta)sin(\theta)f(r\cos(\theta),r\sin(\theta))}{r}=\lim_{r\to 0}rf(r\cos(\theta),r\sin(\theta))cos(\theta)sin(\theta)=0$

The limit is zero because $rf(r\cos(\theta),r\sin(\theta))$ is a function which goes to zero when $r\to 0$ and $\cos(\theta)\sin(\theta)$ is a constant function with respect to $r$, so their product goes to zero without dependence on $\theta$.

Mark
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Your last limit is just

$$\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^2+y^2}}\cdot\lim_{(x,y)\to(0,0)}f(x,y)$$

because the right factor exists by hypothesis, and the left factor is $0$.

symplectomorphic
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  • But that shows me that $g (x, y)$ is differentiable? because by hypothesis I know that $f(x,y)$ is continuous on $(0,0)$ but I do not know if it is differentiable at that point – Ayesca Oct 10 '18 at 17:51
  • I don't understand your comment. All I'm using is the fact that $\lim (f\cdot g) = (\lim f)(\lim g)$ when the two limits exist, and I'm working from the final line of your calculation. If you're asking instead whether your calculation is right or how to justify it, that is a different matter. – symplectomorphic Oct 10 '18 at 17:53