How to prove $\lim\limits_{x\to 0}xf(x)=0$ suppose we know $\lim\limits_{x\to 0}f(x)=1$? How to prove $\lim\limits_{x\to 0}xf(x)=0$ suppose we know $\lim\limits_{x\to 0}f(x)=1$?
So we know that $\forall\epsilon\gt 0,\ \exists\delta\gt 0:0\lt|x|\lt\delta\implies|f(x)-1|\lt\epsilon$
We want to prove $\forall\epsilon\gt 0,\ \exists\delta\gt 0:0\lt|x|\lt\delta\implies|xf(x)|\lt\epsilon$
In my personal opinion, it's equivalent to prove $|xf(x)|\lt |f(x)-1|$. (Not sure)
What steps should I take to do the formal proof?
 A: HINT:
$$|xf(x)|=|x(f(x)-1)+x|\le |x||f(x)-1|+|x|$$
A: All we need
is this theorem:
If
$f$ is bounded
in some neighborhood of $0$,
then
$\lim_{x \to 0} xf(x) = 0
$.
Proof:
Suppose
$f(x) < M$
for
$|x| < c$.
Then,
for $|x| < c$,
$|xf(x)|
< |xM|
$.
To make
$|xM| < \epsilon$,
just make
$|x|
< \dfrac{\epsilon}{M}
$.
(End of proof)
In your case,
since
$\lim_{x \to 0} f(x)
= 1
$,
there is a $\delta$
such that
$|x| < \delta
\implies |f(x)-1| < 1
$.
Note that I am using
$\epsilon = 1$
for the usual limit
specification,
since we just need to get
an upper bound,
any upper bound,
for $f$ in
some neighborhood of $0$.
Therefore
$|f(x)| < 2$
for
$|x| < \delta$.
This gives the
needed neighborhood of $0$
in which $f$ is bounded
(by $2$),
so we can apply
the initial result.
A: If the sequence $f(x)$ is convergent, then after some $n$, it is bounded as well. Let $r$ be a real number such that $|f(x)|<2$ if $|x|<r$. Now, given $\epsilon>0$, note that $|xf(x)|<2|x|$ whenever $|x|<r$. So let $\delta = \frac{1}{2}\min(r,\epsilon/2)$. Note that $|x| < \delta \implies |x| < \epsilon/2 $ and $ |x|<r \implies xf(x)<2x<\epsilon$. This should do it.
