# Proving a sequence is convergent using continuity of function and the fact the derivative is bounded on a given set..

If the function $$f:(0,1] \to \mathbb{R}$$ is differentiable at the semi-open interval $$(0,1]$$ and $$|f'(x)|<1$$ for each $$x \in (0,1]$$. For each $$n \in \mathbb{N}$$ take $$x_{n} \in (0,1]$$ such the sequence $$\lbrace x_{n} \rbrace_{n=1}^{\infty}$$ is convergent to $$0$$. Prove the sequence $$\lbrace f(x_{n}) \rbrace_{n=1}^{\infty}$$ is convergent.

My attempt of proof goes as follow: As the function $$f:(0,1] \to \mathbb{R}$$ is differentiable at the semi-open interval $$(0,1]$$ then $$f$$ is continuous at $$(0,1]$$. More so, as $$|f'(x)|<1$$ for each $$x \in (0,1]$$ then $$f$$ is uniformly continuous at $$(0,1]$$. We want to prove that for each $$\epsilon > 0$$ there is a $$N \in \mathbb{N}$$ such if $$n \geq N$$ then $$|f(x_{n})-f(0)|$$. (I got the intuiton $$\lbrace f(x_{n}) \rbrace_{n=1}^{\infty}$$ is convergent to $$f(0)$$). Because $$\lbrace x_{n} \rbrace_{n=1}^{\infty} \to 0$$ and $$f$$ is continuous in $$(0,1]$$ and each $$x_{n} \in (0,1]$$ by some equivalence of continuity we got $$\lbrace f(x_{n}) \rbrace_{n=1}^{\infty}$$ is convergent to $$f(0)$$. Im I done? What it troubles me is that I didnt use the fact f'(x) is bounded and $$f$$ is uniformly continuous on $$(0,1]$$. Any help with the proof of this exercise will be aprecciated.

• First note that $f(0)$ is not defined, so $|f(x_n)-f(0)|$ does not make any sense. Moreover, as you found out yourself, you didn't use $|f'|<1$ on $(0,1]$. Hint: Show that $f(x_n)$ is a Cauchy sequence, when $(x_n)$ converges to 0. – sranthrop Nov 14 '18 at 21:09
• Thanks @sranthrop ! Your hint was really helpful. I already write my answer below based on your hint. I would aprecciate to check it out and tell if the proof is right at all. – Cos Nov 14 '18 at 22:20

Working through the hint @srnthrop give me I got the following: As $$|f'(x)|<1$$ for every $$x \in (0,1]$$ then f is uniformly continuous in (0,1]. By another side, as $$\lbrace x_{n} \rbrace$$ is convergent to $$0$$ then $$\lbrace x_{n} \rbrace$$ is a Cauchy sequence, this means that for every $$\delta > 0$$ there exist and $$N \in \mathbb{N}$$, if $$n,m \geq N$$ then $$|x_{n}-x_{m}|< \delta$$. But if we take $$\epsilon > 0$$, as we have that $$f$$ is uniformly continuous in $$(0,1]$$, $$|x_{n}-x_{m}|< \delta$$ and $$x_{n}, x_{m} \in (0,1]$$ we conclude $$|f(x_{n})-f(x_{m})|< \epsilon$$. This last inequality means $$\lbrace f(x_{n}) \rbrace$$ is a Cauchy sequence in $$\mathbb{R}$$ implying the sequence $$\lbrace f(x_{n}) \rbrace$$ is convergent.