How to prove $\lim\limits_{x\to 0}xf(x)=0$ suppose we know $\lim\limits_{x\to 0}f(x)=1$?

Question

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?

Answer 1

HINT:

$$|xf(x)|=|x(f(x)-1)+x|\le |x||f(x)-1|+|x|$$

Answer 2

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.

Answer 3

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.