# Limit of composite function approximation

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.

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

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 $$\lim\limits_{x \to 0+}\frac{f(g(x))}{f(x)} = \frac{1}{e} \neq 1$$.
• 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 $1$. Commented Dec 19, 2022 at 8:54
• 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). Commented Dec 19, 2022 at 9:18