Let $I$ be an open interval that contains the point $c$ and suppose that $f$ is a function that is defined on $I$ except possible at the point $c$. Suppose that $|f(y)-f(x)|<\gamma\cdot |y-x|$ for all $x,y \in I-\{c\}$ where $\gamma$ is a positive constant. Prove that the $\displaystyle{\lim_{x \to c}f(x)}$ exists.
My attempt:
Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence in $I-\{c\}$ such that $x_n \xrightarrow[n \to \infty]{}c$. I will use the following theorem:
Theorem. Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence of values in $I/\{c\}$, then $\{f(x_n)\}_{n\in\mathbb{N}}$ is a sequence in the range of $f$, we have $\lim_{n\to \infty}$ $f(x_n)=L$. Then $\displaystyle{\lim_{x \to c}f(x)=L}$.
To show that $(f(x_n))_{n\in\mathbb{N}}$ converges, show that it is a Cauchy sequence (since all Cauchy sequences converge). Choose $\epsilon_1>0$. Since $x_n$ converges to $c$, it is a Cauchy sequence. Let $\epsilon_1=\epsilon_2/\gamma$ and so there exists an $N$ in the the natural numbers such that $|x_n-x_m|\leq \epsilon_2$ if $n,m\geq N$. Then $|f(x_n)-f(x_m)|<\gamma\cdot|x_n-x_m|\leq \epsilon_1$. Therefore by that theorem, $\displaystyle{\lim_{x \to c}f(x)}$ exists.