if a function $f$ is decreasing and the limit $\lim\limits_{x\to \infty} f(x)$ exists, then Given that function $f$ is decreasing and the limit $\lim\limits_{x\to \infty} f(x)$ exists
How can I prove $$\lim\limits_{x\to \infty} x\left(f(x)-f(x+1)\right)$$ exists?
I applied monotone convergence theorem, but there is no way...

This is false. Counterexample:
Define $f(x)$ this way, for each positive integer $n$ then for all $n^{2}<x\leq (n+1)^{2}$ then $f(x)=-\sum\limits_{i=1}^{n}\frac{1}{i^{2}}$. Clearly $f(x)$ is bounded below (bounded by $-\frac{\pi^{2}}{6}$) and monotone decreasing.
Thus whenever $x=(n+1)^{2}$ then $x(f(x+1)-f(x))=x(-\sum\limits_{i=1}^{n+1}\frac{1}{i^{2}}-(-\sum\limits_{i=1}^{n}\frac{1}{i^{2}}))=(n+1)^{2}\frac{-1}{(n+1)^{2}}=-1$. But most of the time, (whenever $x=(n+1)^{2}-1$ for example) $x(f(x+1)-f(x))=0$ instead. Hence this oscillate, so limit does not exist.
