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Apostol' Calculus gives $o$-notation as follows:

Assume $g(x) \neq 0$ for all $x \neq a$ in some interval containing $a$. The notation $$ f(x) = o(g(x)) \quad \text{as}\ x\to a $$ means that $$\lim_{x\to a}\frac{f(x)}{g(x)} = 0.$$

how to prove $o(f(x)) + o(g(x)) = o(f(x) + g(x))$ by using this definition?

Edit:

Now I can only prove an easier case, if $F(x)\in o(f(x))$, $G(x) \in o(g(x))$, for any $c > 0$, we have $|F(x)| < c|f(x)|$, $|G(x)| < c|g(x)|$, then $\lim_{x\to a}\frac{|F(x)| +|G(x)|}{|f(x)| + |g(x)|} = 0$, also $\lim_{x\to a}\frac{|F(x)+ G(x)|}{|f(x)| + |g(x)|} = 0$, $$ o(f(x)) + o(g(x)) = o(|f(x)| + |g(x)|) = o(\max\{|f(x)|, |g(x)|\}). $$

It seems that we should restrict $f(x)$ and $g(x)$ in $o()$ to be positive, or there are some counterexamples in $o(f(x)) + o(g(x)) = o(f(x) + g(x))$.

yanpengl
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  • Please [edit] the question to show us what you tried and where you are stuck. – Ethan Bolker Dec 21 '21 at 15:39
  • Let $F,,G$ denote the functions added on the LHS. Hint: can you prove $$F/f\to0,,G/g\to0\implies(F+G)/(f+g)\to0$$with one or more suitable inequalities? – J.G. Dec 21 '21 at 15:41
  • @J.G. from the definition of the little-o notation, I can only prove a weaker conclusion: $o(f(x)) + o(g(x)) = o(|f(x)| + |g(x)|)$. – yanpengl Dec 22 '21 at 06:24
  • @EthanBolker I edited. – yanpengl Dec 22 '21 at 06:51
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    As stated this is false, if the denominators are allowed to be negative. Take for example $F(x)=G(x)=x$, $f(x) = -x^2$, and $g(x) = x^2+1$ for the limit $x\to\infty$. (For $u\to0$ use the same example with $x$ replaced by $\frac1u$.) The absolute value version is probably the best you can do (and I've always seen this notation defined with the restriction that the denominators must in fact be positive). – Greg Martin Dec 22 '21 at 06:59
  • @GregMartin yes, since $o(g(x)) = o(|g(x)|)$, the definition of little-oh should be defined for positive denominators, which is better than the one given by this book. – yanpengl Dec 22 '21 at 07:30
  • If you restrict the denominators to be positive, have not you yourself proved your problem in the question? On the other hand, if they are allowed to be negative, then a counter-example is provided by @GregMartin – joy Dec 22 '21 at 10:22
  • @joy yes, if restricting the denominator to be positive in the definition(to some extend, modify the definition in this way), $o(f(x)) + o(g(x)) = o(f(x) + g(x))$ is true. if no this restriction, this proposition is false, but $o(f(x)) + o(g(x)) = o(|f(x)| + |g(x)|)$ is true. – yanpengl Dec 22 '21 at 11:19

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