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?
 A: 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 $a_n<x\le a_{n+1}$ and hence by the MVT $|F(x)-F(a_{n+1})|=|F'(\xi)|\cdot|a-a_{n+1}|<\frac{\epsilon}{2B}\cdot B=\frac\epsilon2$ with $\xi\in (x,a_{n+1})$. As also $|F(a_{n+1}-l|<\frac\epsilon2$ we conclide $|F(x)-l|<\epsilon$.
We conclude that  for all $x>M$ we have $|F(x)-l|<\epsilon$. In summary, $F(x)\to l$.

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).
A: Suppose $a_{n+1}-a_n \le B$ for all $n$.
Let $\epsilon>0$, and choose $K$ such that for all $x \ge K$ we have
$|F'(x)| \le {1 \over 2B} \epsilon$.
Now choose $N$ such that for $n \ge N$ we have $a_n \ge K$ and $|F(a_n) -L| < {1 \over 2 } \epsilon$. Let $K' = A_N$.
Now suppose $x \ge K'$, then $x \in [a_n,a_{n+1}]$ for some $a_n \ge K$ and
so $|F(x)-F(a_n)| \le { 1\over 2B} \epsilon |x-a_n| \le { 1\over 2} \epsilon $.
Then $|F(x)-L| \le |F(x)-F(a_n)| + |F(a_n)-L| < \epsilon$.
A: Hint: Suppose $0<a_n-a_{n-1}<L$ for all $n$. For given $\varepsilon>0$, $n$ large enough and $x$ between $a_{n-1},a_n$ we have $|F(x)-F(a_{n-1})|=|F'(\theta)(x-a_{n-1})|\leq |F'(\theta)(a_n-a_{n-1})|<\varepsilon L$.
