# Thread: g'(0) exists s.t. lg(x)l <or= (x^2) proof

1. ## g'(0) exists s.t. lg(x)l <or= (x^2) proof

this question has two parts. i have tried to answer part a) but I don't think my proofs are very good. I'm practicing for an exam so if someone could explain the concepts I am supposed to have gotten from the question, I would very much appreciate it.

note: a hint to part a) is that the previous question (which i solved) was to prove f is differentiable at 0 if f(x)=(x^2) for rational x and f(x)=0 for irrational x. I don't think I fully understand the connection to this problem--again, any conceptual help would be much appreciated.

a) let g be a function such that lg(x)l <or= (x^2) for all x. Prove that g is differentiable at 0.

b) this result can be generalized if (x^2) is replaced by lf(x)l where f has what property?

2. Originally Posted by 234578
this question has two parts. i have tried to answer part a) but I don't think my proofs are very good. I'm practicing for an exam so if someone could explain the concepts I am supposed to have gotten from the question, I would very much appreciate it.

note: a hint to part a) is that the previous question (which i solved) was to prove f is differentiable at 0 if f(x)=(x^2) for rational x and f(x)=0 for irrational x. I don't think I fully understand the connection to this problem--again, any conceptual help would be much appreciated.

a) let g be a function such that lg(x)l <or= (x^2) for all x. Prove that g is differentiable at 0.

b) this result can be generalized if (x^2) is replaced by lf(x)l where f has what property?
Hint:
Spoiler:

a) $|g(x)|\leqslant x^2\implies -x^2\leqslant g(x)\leqslant x^2\implies -x\leqslant\frac{g(x)}{x}\leqslant x$

b) $\frac{f(x)}{x}\underset{x\to0}{\sim}0$
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