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The function is as follows:


and I want to calculate the following limit:


The reason I'm having trouble with this one is because the limit doesn't seem to be $0$ but $y_0^2$. Because of that, I need 2 functions to compare $f$ to, instead of one. The greater one I found like this:

$\ln(1+x^2y^2)\leq x^2y^2$

$\frac{\ln(1+x^2y^2)}{x^2}\leq y^2$

so (if I'm correct) the limit is definitely lower or equal to $y_0^2$. But I can't find the function to be my upper bound that also converges to that value.

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sorry i had not read question and problem exactly,i have updated – dato datuashvili Feb 3 '14 at 9:33
up vote 2 down vote accepted

$$\ln(1+x) \sim x$$ if $x \to 0.$

So you have $$\frac{\ln(1+x^2y^2)}{x^2} \sim y^2$$

$$\lim_{x, y \to 0, y_0} y^2 = y_0^2$$

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Are you saying that my upper bound is enough to prove the limit? Could you offer some more explanation? – Luka Horvat Feb 3 '14 at 9:34
@Darwin i have updated my answer – dato datuashvili Feb 3 '14 at 9:35
@Darwin usually you need to find upper bounds to simplify the limit. But it is always true that you can replace a function in a limit with it's taylor approximation. Doing so we know that the limit is equal to $y_0^2$, so no need to do anything else ;-) – Ant Feb 3 '14 at 9:41

You can use the fact that $$\ln(1+u)=u + o(u)$$ when $u$ converges to $0$.

Here you can apply this to $u=x^2y^2$, to get that $$f(x,y)=\frac{x^2y^2+o(x^2y^2)}{x^2}=y^2+o(y^2)$$

The last $o(y^2)$ converges to $0$ when taking the limit, so you are done.

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Why is $\frac{o(x^2y^2)}{x^2}=o(y^2)$? – Luka Horvat Feb 3 '14 at 9:39
$o(sth)$ means here "a function $h(x,y)$ such that $h(x,y)/sth$ converges to $0$ when $x^2y^2\to 0$". From this definition, it should be clear that you can do this. – zozoens Feb 3 '14 at 9:43

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