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I'd be interested to know when, if $$f(x,y)=g(x,y)+O(x^n)$$ we have that $$\lim_{y\rightarrow c}=\lim_{y\rightarrow c}g(x,y)+O(x^n).$$ Are there conditions of $f$ and/or $g$ that make sure that this is satisfied?

In particular, assume that we have a function $f(x,y)$ and that we Taylor expand it as a function of $x$ around $a$: $$f(a+x,y)=f(a,y)+\frac{\partial f}{\partial x}(a,y)\cdot x+\frac{\partial^2 f}{\partial x^2}(a,y)\cdot \frac{x^2}{2}+\cdots+O(x^n).$$

I wish to study the behaviour of $f(x,y)$ as $y$ tends to a value on the boundary of the domain of $f(x,y)$, say $y\rightarrow c$. I therefore wonder under what conditions on $f$ $$\lim_{y\rightarrow c}f(a+x,y)=\lim_{y\rightarrow c}f(a,y)+\lim_{y\rightarrow c}\frac{\partial f}{\partial x}(a,y)\cdot x+\lim_{y\rightarrow c}\frac{\partial^2 f}{\partial x^2}(a,y)\cdot \frac{x^2}{2}+\cdots+O(x^n).$$ In other words, when is the $O(x^n)$ term still $O(x^n)$ as $y\rightarrow c$?

I've been hoping to find some smoothness constraints on $f$ that would guarantee that the expansion still is $O(x^n)$, but so far I haven't been able to find any.

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Not homework, btw. –  MånsT Jul 25 '12 at 8:38
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I think you want $O(x^n)$ to be uniform with respect to $y$ (in a neighborhood of $c$). More precisely, you need to bound the remainder by $C|x|^n$ with $C$ independent of $y$. The Lagrange form of the remainder is probably most likely to help. Using it, you will need to estimate the $n$th derivative of $f$ with respect to $x$ from above; this is possible, for example, if the derivative is jointly continuous in the variables $x$ and $y$.

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Thanks! I can't believe that I didn't think of the Lagrange form of the remainder... I guess that $C$ can depend on $y$ if it is well-behaved as $y\rightarrow c$? –  MånsT Jul 26 '12 at 15:13
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@MånsT Yes, it can be put this way too: the remainder estimate of the form $\le C(y)|x|^n$ is okay as long as $C(y)$ stays bounded as $y\to c$. –  user31373 Jul 26 '12 at 16:01
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