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. Sinceas
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
- False.
Considerfor
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.