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# Prove or disprove given statements for functions such that f(x) = o(g(x))

Let and be functions, both differentiable in a neighborhood of 0, with and such that Prove or disprove each the following statements.

1. as .
2. as .

1. True.
Proof. Since as we know by the definition of that Thus, for every there exists a such that So, for we have The final line follows since by hypothesis. Therefore, Hence, By definition, we then have 2. False.
Consider for and for . Then, for , For we have .

Next, Since we have as . However, since does not exist.

1. hteica says:

I think part (b) could use a more simpler counterexample. For example, let $f(x)=x$ and $g(x)=1$ defined for all real numbers. Then $f$ and $g$ both have derivatives in some interval containing 0, and clearly $g$ is always positive, also $f(x)=o(g(x))$ as $x\rightarrow 0$. But we can see that $1=f^{\prime}(x)=o(g^{\prime}(x))=o(0)$ is not true, since o(0) is undefined.

• Artem says:

This counterexample is actually quite interesting :) I wonder if it is legit to conclude the case from the non-existence of o(0). If you have a question somewhere on stack, feel free to share for everyone :)

2. Zerong Xi says:

There is a fault in your last step. lim(x->0) sin(1/x)/(1/x) = 0 instead of 1.