Limit of Product exists if the limit of the function decays to zero

Question

If $f:\mathbb{R}\to \mathbb{R}$ such that $\lim\limits_{x \to \infty}xf(x)=L$. Then $\lim\limits_{x \to \infty}f(x)=0$.

My proof is as follows:

Let $g(x)=xf(x)$. Fix an $\epsilon>0$, then by definition, there exists a $\delta>0$ such that $$x>\delta \implies |g(x)-L|<\epsilon$$ Then $$|f(x)|=\frac{|g(x)|}{|x|}\le \frac{L+\epsilon}{\delta}$$ claiming the result.

Are there any flaws in my argument?

Answer 1

Unfortunately,

$$|f(x)|\le\frac{L+\epsilon}\delta$$ is not conclusive. Nothing says that $\delta$ is even large.

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A simple proof is

$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\left(xf(x)\cdot\frac1x\right)=\lim_{x\to\infty}(xf(x))\cdot\lim_{x\to\infty}\frac1x=L\cdot0.$$

Answer 2

$lim_{x\rightarrow \infty} f(x)=L\in \mathbb{R}$ if and only if

for all $\epsilon>0$ there exists $\delta>0$ such that for all $x>\delta$, $|f(x)-L|<\epsilon$

Let us prove your problem.

Supposing $lim_{x\rightarrow \infty}xf(x)=L$ , we wish to show that $lim_{x\rightarrow \infty}f(x)=0$.

To this end, let $\epsilon>0$ be arbitrary. As $lim_{x\rightarrow \infty}xf(x)=L$, there exists $\delta_1>0$ such that for all $x>\delta_1$, $|xf(x)-L|<\epsilon$.

Hence, for $\delta>max(\delta_1,\frac{\epsilon+|L|}{\epsilon})$, whenever $x>\delta$, we get

$|f(x)|=|\frac{xf(x)}{x}|$=$\frac{1}{x}|xf(x)-L+L|\leq \frac{1}{x}|xf(x)-L|+\frac{1}{x}|L|\leq \frac{1}{\delta}(\epsilon+|L|)\leq \epsilon$