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I am interested to find the limit of the following functions,

$$\lim_{(x,y) \to (0,0)} \frac{\ln(1+xy)}{\sqrt{x^2+y^2}}$$

$$\lim_{(x,y) \to (0,0)} \frac{x^2 y}{x^4+y^2}$$

If the limit is not existing then what should be the reason of its not existance.

Martin Sleziak
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Kiran
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  • There is another post about the second limit: http://math.stackexchange.com/questions/1476353/help-with-lim-x-y-0-0-fracx2yx4y2 – Martin Sleziak Oct 12 '15 at 14:18
  • Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. – Martin Sleziak Oct 12 '15 at 14:32

2 Answers2

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Along the path $y=x^2$ shouldn't the second limit equal $\frac{1}{2}$? Since along the x-axis or the y-axis the limit equals zero, wouldn't that mean that the second limit doesn't exist?

Gondim
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By using the polar transformation, we have: $$ \lim_{(x,y)\rightarrow(0,0)}\frac{\ln(1+xy)}{\sqrt{x^2+y^2}}=\lim_{r\rightarrow 0}\frac{\ln(1+r^2\sin(\theta)\cos(\theta))}{r}=\lim_{r\rightarrow 0}\frac{r^2\sin(\theta)\cos(\theta)}{r}=0 $$ Also: $$ \lim_{(x,y)\rightarrow(0,0)}\frac{x^2y}{x^4+y^2}=\lim_{r\rightarrow 0}\frac{r^3\cos(\theta)^2\sin(\theta)}{r^4\cos(\theta)^4+r^2\sin(\theta)^2}=0 $$

Lion
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