Big O, little O and equivalent functions.

Question

Suppose that $f,g: [1,\infty) \rightarrow [1,\infty)$ are two positive increasing monotone functions such that $\lim_{x\rightarrow +\infty} f(x) = \lim_{x\rightarrow +\infty} g(x) = + \infty$. Which of these are true? $$ \begin{array}{l} A)\ If\ f(x) \sim g(x)\ then\ g(x) \sim f(x).\\ B)\ If\ f(x) = o(g(x))\ then\ g(x) = o(f(x)).\\ C)\ If\ f(x) = O(g(x))\ then\ g(x) = O(f(x)).\\ D)\ If\ f(x) \sim g(x)\ then\ (f(x))^2 \sim (g(x))^2. \\ E)\ If\ f(x) = o(g(x))\ then\ (f(x))^2 = o((g(x))^2). \\ F)\ If\ f(x) = O(g(x))\ then\ (f(x))^2 = O((g(x))^2). \\ G)\ If\ f(x) \sim g(x)\ then\ \ln(f(x)) \sim \ln(g(x)). \\ H)\ If\ f(x) = o(g(x))\ then\ \ln(f(x)) = o(\ln(g(x))). \\ I)\ If\ f(x) = O(g(x))\ then\ \ln(f(x)) = O(\ln(g(x))). \\ J)\ If\ f(x) \sim g(x)\ then\ 2^{f(x)} \sim 2^{g(x)}. \\ K)\ If\ f(x) = o(g(x))\ then\ 2^{f(x)} = o(2^{g(x)}). \\ L)\ If\ f(x) = O(g(x))\ then\ 2^{f(x)} = O(2^{g(x)}). \\ \end{array} $$

Well, it is easy to see that A is true and $B,C$ are false, but how do you operate with these in the rest of the cases?

Answer 1

D, E and F are true because of the definition of $\sim$, $O$ and $o$ and product operation on limits.

G is true because ${\ln {f\over g}\over \ln g} = {\ln f\over \ln g} - 1$ converges to 0 (since $\ln g$ diverges to $+\infty$)

H is not true (for instance take $f = x$ and $g = x^2$)

I is true because if $f = O(g)$ then there exists M such that $f \le M g$ and since $\ln$ is increasing, you get $\ln f \le \ln M + \ln g$, and $\ln g$ diverges to $+\infty$

J is not true (for instance $f = x+1$ and $g = x$)

K is true because ${f\over g} -1 \rightarrow -1$ thus $f-g \rightarrow -\infty$ and therefore $2^{f-g} \rightarrow 0$

L is not true ($f = 2x$ and $g = x$)

Answer 2

since $f(x)>0$ and $g(x)>0$, you can use the following definition of $f \sim g$: $\lim_{x \to \infty} \frac{f(x)}{g(x)}=1$
D) $\lim_{x \to \infty} (\frac{f(x)}{g(x)})^2=1^2$ so that $\lim_{x \to \infty} \frac{f^2(x)}{g^2(x)})=1$ so that $f \sim g$
etc.
You can use the following definition for $f=O(g)$ : $\lim_{x \to \infty} \frac{f(x)}{g(x)}<+\infty$ (so for example F is true)
You can use the following definition for $f=o(g)$ : $\lim_{x \to \infty} \frac{f(x)}{g(x)}=0$ (so for example E is true)

Answer 3

These become a lot clearer if you write out the definitions. I will give two examples:

D) The definition of $f(x) \sim g(x)$ is that $\lim_{x \to \infty} \frac{f(x)}{g(x)} = 1$. It now follows from the limit theorems that $\lim_{x \to \infty} (\frac{f(x)}{g(x)})^2 = (\lim_{x \to \infty} \frac{f(x)}{g(x)})^2 = 1$, and therefore $(f(x))^2 \sim (g(x))^2$.

E) The definition of $f(x) \in o(g(x))$ is that $\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0$. So again we have that $\lim_{x \to \infty} (\frac{f(x)}{g(x)})^2$ = $(\lim_{x \to \infty} \frac{f(x)}{g(x)})^2 = 0$. Conclusion: $f(x))^2 \in o((g(x))^2$.

The other proofs go similarly.