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Given a function $f:\mathbb R^2 \rightarrow \mathbb R$ is continuous and has a limit at $p=\infty$, $\lim_{x\rightarrow p}f(x,y)=b(y)$ and a function $g:\mathbb R \rightarrow \mathbb R$ is continuous and has a limit at $p$, $\lim_{x\rightarrow p}g(x)=c$, is the following statement true? $$\lim_{x\rightarrow p}f(x,g(x))=\lim_{x\rightarrow p}f(x,c)=b(c)$$

Igor
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    Yes: it is true. It is a direct consequence of continuity of $f,g$. – Crostul Jul 30 '15 at 22:18
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    If the functions are continuous and defined everywhere then we do not need to talk about limits - it is just the functional value $f(p,g(p))$. – A.Γ. Jul 30 '15 at 22:22
  • In general, there is no such rule for the limits. It is not recommended to take partial (sequential) limits with respect to some $x$ while keeping other as $x$, it may easily give a wrong result, for example, $\lim_{x\to 0}x\cdot\frac{1}{x}=\lim_{x\to 0}0\cdot\frac{1}{x}=0$ is wrong as you may see. – A.Γ. Jul 30 '15 at 22:32
  • Yes, I know that in general the chain rule does not apply to limits. I want to be sure that I am correctly applying it in this specific case I outline. In the example you give, $lim_{x \rightarrow 0} \frac 1 x$ does not exist so it doesn't fit my example. – Igor Jul 30 '15 at 22:40
  • @A.G. Note that the question says that $p = \infty$. –  Jul 30 '15 at 22:53
  • @Ezueneok oh, really? I've missed it. Then it is not true. – A.Γ. Jul 30 '15 at 23:07
  • Is there a notion of "continuous at infinity" that will make this statement true? Or any additional restrictions on $f$ and $g$? – Igor Jul 30 '15 at 23:14
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    I guess your $b(y)$ must be continuous too. What you basically do is the switching of limit order: you need to know when $\lim_{x\to p}\lim_{y\to c}f(x,y)=\lim_{y\to c}\lim_{x\to p}f(x,y)$. As far as I remember, one of the limits should be uniform, it is sufficient, but check it. – A.Γ. Jul 30 '15 at 23:24
  • Check out here. – A.Γ. Jul 30 '15 at 23:38

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For $p=\infty$ it is not true. A counterexample: $$ f(x,y)=\frac{1}{x^2|y|+1}\to b(y)=\left\{ \begin{array}{ll} 0, & \text{if } y\ne 0,\\ 1, & \text{if } y= 0. \end{array} \right. $$ $$ g(x)=\frac{1}{x^2+1}\to c=0. $$ So we have $f(x,c)=f(x,0)=1$, but $$ f(x,g(x))=\frac{1}{\frac{x^2}{x^2+1}+1}\to \frac12. $$

A.Γ.
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