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Suppose I have the following limit $$\lim_{x\to0} f(g(x))$$ where $f,g:\mathbb{R}\to \mathbb{R}$, and suppose that $g(x)\sim x$ for $x\to0$ in the sense that $\lim_{x\to0} \frac{g(x)}x =1$. Could I say that $f(g(x))\sim f(x)$? And so that $$\lim_{x\to0} f(g(x)) = \lim_{x\to0} f(g(x))?$$ I should prove that $$\lim_{x\to0}\frac{f(g(x))}{f(x)} = 1$$ but I don't know how.

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  • $\begingroup$ Do you know anything about $f$? Without it, $f(g(x))$ can be pretty much arbitrary function. $\endgroup$
    – mihaild
    Commented Dec 12, 2022 at 13:04
  • $\begingroup$ Yes, of course both $f$ and $g$ are defined on the real line, continuous in a neighbourhood of $0$ $\endgroup$
    – aleio1
    Commented Dec 12, 2022 at 13:08

1 Answer 1

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Not necessary. $f(x)$ can go to zero very fast, so changing it's argument even by $o(x)$ multiplies it by separated from $1$ constant.

For example, take $f(x) = \exp(-1/|x|)$ and $g(x) = x + x^2$.

Then, for $x > 0$, we have

$$\frac{f(g(x))}{f(x)} = \exp\left(\frac{1}{x + x^2} - \frac{1}{x}\right) = \exp\left(\frac{-x^2}{x^2 + x^3}\right) = \exp\left(-1 + \frac{x}{1 + x}\right) $$

And so $\lim\limits_{x \to 0+}\frac{f(g(x))}{f(x)} = \frac{1}{e} \neq 1$.

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  • $\begingroup$ Ok, maybe I put the question in the wrong way. I meant, is it true that if $g(x)\sim x$, then $f(g(x))$ has the same limit of $f(x)$? I suppose that if the two limits are both finite and not $0$ then it holds that the ratio $\frac{f(g(x))}{f(x)}$ approaches to $1$. $\endgroup$
    – aleio1
    Commented Dec 19, 2022 at 8:54
  • $\begingroup$ If $f$ and $g$ are both continuous and $g(0) = 0$, then $\lim_{x \to 0} f(g(x)) = \lim_{x \to 0} f(x) = f(0)$ (by continuity of composition), we don't need any assumptions on growth rate of $g$. If, moreover, $f(0) \neq 0$, this implies $\frac{f(g(x))}{f(x)} \to 0$ (by continuity of quotient). $\endgroup$
    – mihaild
    Commented Dec 19, 2022 at 9:18

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