Prove that $\lim_{x\to\infty} f(x) = 0$ if $\lim_{x\to\infty} xf(x) = L$ Question:
Let $f: (a,\infty) \to \mathbb{R}$ be such that $\lim_{x\to\infty} xf(x) = L$ where $L\in \mathbb{R} $. 
Prove that $\lim_{x\to\infty} f(x) = 0$.
Attempt:
I see that in order for $\lim_{x\to\infty} xf(x) = L$, either $f(x)$ must be bounded and strictly decreasing or $x \to 0$ for the limit to be $L$. Ruling out the latter because that's not what I'm trying to prove, I looked at a bounded monotone decreasing $f(x)$.
Splitting the limit into two:
$\lim_{x\to\infty} xf(x) = \lim_{x\to\infty} f(x) * \lim_{x\to\infty} x = L$
The second term in the middle goes to infinity as $x \to \infty $, which means that to counteract the $\infty$, the $f(x)$ must decreasing faster than linearly and must go to zero.
I'm wondering if this makes sense. Thank you!
 A: There is no reason to assume that $f$ is monotone decreasing, consider for example $f(x)=(\sin x/x)^2$ for $x\geqslant1$, which is neither decreasing nor increasing.
To prove the result, note that, if $xf(x)$ has a limit when $x\to+\infty$ then $xf(x)$ is bounded for $x$ large enough. Let us assume only that $xf(x)$ is bounded for $x$ large enough, that is, that there exists $C$ and $x_0$ such that $|xf(x)|\leqslant C$ for every $x\geqslant x_0$. 
Then $-C/x\leqslant f(x)\leqslant C/x$ for every $x\geqslant x_0$ and both the LHS and the RHS converge to $0$. By the squeeze theorem, $f(x)\to0$ when $x\to+\infty$.
A: Let $\epsilon>0$. Then there exists some $M>1$ such that if $x \ge M$, then $|L-x f(x)|  < \frac{1}{2}\epsilon$.
Dividing across by $x$ gives $|\frac{L}{x} - f(x)| < \frac{\epsilon}{2x} \le \frac{1}{2} \epsilon$.
We have $|f(x)| \le |\frac{L}{x} - f(x)| + |\frac{L}{x}|$. Now if $x \ge M' = \max(M, \frac{2|L|}{\epsilon}+1)$ we have $|f(x)| < \epsilon$.
Hence $f(x) \to 0$.
A: $Proof.$ Let $f:(a, \infty) \rightarrow \mathbb{R}$ be such that $\lim_{x \rightarrow \infty} xf(x) = L$ where $L \in \mathbb{R}$, we will show that $\lim_{x \rightarrow \infty} f(x) = 0$.  $\forall\ \varepsilon > 0.$ Choose $K = (|L| + 1)/ \varepsilon$. $\forall\ x > K$ Since $xf(x) \rightarrow L$ as $x \rightarrow \infty$, $$|xf(x) - L| < 1\ $$ i.e. $$|xf(x)| \le |L| + 1$$
So $$|f(x) - 0| = \left| \frac{xf(x)}{x} \right| \le \frac{|L|+1}{|x|} < \frac{|L|+1}{K} = \frac{|L|+1}{(|L| + 1)/ \varepsilon} = \varepsilon.$$ Therefore $lim_{x \rightarrow \infty} f(x) = 0.$
