Let and be functions, both differentiable in a neighborhood of 0, with and such that

Prove or disprove each the following statements.

- as .
- as .

- True.

*Proof.*Since as we know by the definition of thatThus, 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

- False.

Consider for and for . Then, for ,For we have .

Next,

Since we have as . However, since

does not exist.

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.

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 :)

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