Question: Let $f:(0,1) \to \mathbb{R}$ be a differentiable function such that $|f'(x)| \leq 5$, for all $x \in (0,1)$. Show that the sequence $\{f(\frac{1}{n+1})\}$ converges in $\mathbb{R}$.
Attempt:
Consider any $a, b$ such that $[a,b] \subset (0,1)$. By LMVT $|\frac{f(b)-f(a)}{b-a}| \leq 5$ $\implies |f(b)-f(a)|\leq 5|b-a|$.
The sequence $\{\frac{1}{n+1}$} converges. Hence, by Cauchy's General Principle, $\forall \epsilon >0$, $\exists \ N>0$ such that $|\frac{1}{n+1}-\frac{1}{m+1}| < \epsilon/5$ for every $m, n >N$.
Hence, $ |f(\frac{1}{n+1})-f(\frac{1}{m+1})|\leq 5|\frac{1}{n+1}-\frac{1}{m+1}|< \epsilon$, whenever $m, n >N$.
PS: Please do check the solution thoroughly. Kindly correct any "false assumptions", "weak -wordings", "misplaced predicates" etc. A little help goes a long way. I am struggling with proper proof writing right now.
An other approach.
$$(\forall x\in (0,1)) \; |f'(x)|\le 5 \implies $$
$$f \text{ is Uniformly continuous at } (0,1) \implies $$
$$\lim_{x\to 0^+}f(x) \text{ exists in } \Bbb R \implies$$
$$\lim_{n\to +\infty}f(\frac{1}{n+1})\in \Bbb R.$$
PS
I know that tomorrow my total points will decrease with NO reason. should i change my name.