calculus - prove that if $lim limits_{n to infty}F( a_n)=ell$, then $lim limits_{x to infty}F( x)=ell$

Let $a_n$ be a strictly increasing sequence of positive real numbers ($a_n>0$) and $\lim \limits_{n \to \infty} a_n=\infty.$ Suppose that the sequence $(a_{n+1}-a_n)_n$ is bounded. Let $F:\mathbb{R}^+\rightarrow \mathbb R$ be a differentiable function. Ssuppose further that $\lim \limits_{x \to \infty}F'(x)=0$ and $\lim \limits_{n \to \infty}F( a_n)=\ell$. (Note that $\ell$ is a real number). Prove that: $$\lim \limits_{x \to \infty}F( x)=\ell$$ I'm stuck, I've tried to use the MVT but it gets me nowhere! can anyone help me solve this problem?

Answer

Let $B$ be an uper bound for $|a_{n+1}-a_n|$.

Let $\epsilon>0$.

Pick $M\in \Bbb R$ large enough so that $M>a_1$ and such that $|F'(x)|<\frac{\epsilon}{2B}$ for all $x>M$ (possible because $F'(x)\to 0$) and $|F(a_n)-l|<\frac\epsilon2$ for all $n$ with $a_n>M$ (possible because $F(a_n)\to l$).
Let $x>M$. Tnen there exists $n$ with $a_nWe conclude that for all $x>M$ we have $|F(x)-l|<\epsilon$. In summary, $F(x)\to l$.

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Some of the given conditions are indeed necessary:

• Consider $F(x)=x$ and $a_n=2-\frac1n$ and $l=2$. Here $a_n\not\to \infty$



• Consider $F(x)=\sin x$ and $a_n=n\pi$ and $l=0$. Here $F'(x)\not\to 0$ and the conclusion does not hold.


• Consider $F(x)=\sin \sqrt x$ and $a_n=n^2\pi$ and $l=0$. Here $|a_n-a_{n+1}|$ is not bounded and the conclusion fails.

However, the claim remains valid also if some (necessarily only finitely many) of the $a_n$ are $\le 0$, or if $a_n$ fails to be strictly increasing. (A closer look at the proof above reveals tha twe did not really use thiese properties, though they may simplify the argumentation).