Consider two real functions $f$ and $g$. I write $f \sim_{x \to x_0 } g$ when $$\lim_{x \to x_0} \frac{f}{g}=1 $$ ($x_0$ can also be infinite)
Suppose that I know that $f \sim_{x \to x_0 } g$. Now take another real function $h$. Under what conditions can I say the following?
$$f \sim_{x \to x_0 } g \implies h \circ f \sim_{x \to x_0 } h \circ g$$
I did not find any theorem on this implication. Is there any one?
If not, then could you suggest me the most common cases in which the implication is true? I know the following ones but I would like to know if there are others.
- $f \sim_{x \to x_0 } g \implies \mathrm{ln}f \sim_{x \to x_0 } \mathrm{ln}g $
- $f \sim_{x \to x_0 } g \implies f^\alpha \sim_{x \to x_0 } g^a \,\,\, \mathrm{with} \, \,\alpha \,\in \mathbb{R} $
- $f \sim_{x \to x_0 } g \implies f^{1/x} \sim_{x \to x_0 } g^{1/x} $
- $f \sim_{x \to x_0 } g \, \, \mathrm{and} \,\, \lim_{x \to x_0} f-g=0 \implies e^f \sim_{x \to x_0 } e^g $
Okay, you need much more technical assumptions. An idea how to approach it:
Assume $\lim_{x\to x_0} f(x)$ exists. Then the only interesting cases are only if the limit is $0$ or $\pm \infty$. For now, consider the case $f(x)\to 0$ and $h(x)\to h(0)=0$ for $x\to 0$.
Assume additionally that $h$ is differentiable at $0$ with nonzero derivative. Then, from $f(x)/g(x)\to 1$ follows that $g(x)\to 0$ for $x\to 0$ and we have $$ \frac{h(f(x))}{h(g(x))} = \frac{h(f(x))}{f(x)} \frac{g(x)}{h(g(x))} \frac{f(x)}{g(x)} \to \frac{h'(0)}{ h'(0) }\cdot 1 =1.$$
In case of $h(x)\to \pm\infty$ and differentiable $f,g,h$ you could apply l'Hopital. The same goes for $f(x)\to \pm \infty$. It becomes much more complicated, if the limits do not exist.