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.

  • $\begingroup$ What does 'abs(f(y)-f(x))0' mean? What is e1? $\endgroup$ – copper.hat Nov 21 '12 at 17:42
  • $\begingroup$ I began editing your post but then I got tired. Please go to the FAQ section to check how to use LaTeX to properly write mathematics in this site. $\endgroup$ – DonAntonio Nov 21 '12 at 17:43
  • 1
    $\begingroup$ @user23793 Please note editing your question. Always use LaTeX to write code in math mode. See some LaTeX tutorial on the internet. $\endgroup$ – MathOverview Nov 21 '12 at 17:57
  • $\begingroup$ Without going into details (my eyes hurt!), your proof is correct basically. $\endgroup$ – DonAntonio Nov 21 '12 at 17:59
  • $\begingroup$ You just need uniform continuity. See, eg, math.stackexchange.com/questions/241825/… $\endgroup$ – copper.hat Nov 21 '12 at 18:07

You are almost finished. You proved that if $(x_n)_{n \in \mathbb{N}}$ is a sequence such that $x_n \to c$ then $f(x_n)$ is a Cauchy sequence $\Rightarrow$ a convergent sequence.

How do you know that for every $(x_n)_{n \in \mathbb{N}}$ s.t. $x_n \xrightarrow[n \to \infty]{}c$ the $\displaystyle{\lim_{n \to \infty} f(x_n)}$ is the same?


Use reductio ad absurdum: Suppose $(x_n)_{n \in \mathbb{N}}$ , $(y_n)_{n \in \mathbb{N}}$ are s.t. $x_n \xrightarrow[n \to \infty]{}c$ , $y_n \xrightarrow[n \to \infty]{}c$ , $\displaystyle{\lim_{n \to \infty} f(x_n)= L_1, \ \lim_{n \to \infty} f(y_n) = L_2}$ and $L_1\neq L_2$. Then find a sequence $(z_n)_{n \in \mathbb{N}}$ which contain $(x_n)_{n \in \mathbb{N}}$ and $(y_n)_{n \in \mathbb{N}}$ as subsequences and $z_n \xrightarrow[n \to \infty]{}c$. Are $\displaystyle{\lim_{n \to \infty} f(z_n)}$ and $L_1$ equal? What about $\displaystyle{\lim_{n \to \infty} f(z_n)}$ and $L_2$?

  • $\begingroup$ Since the values in the domain of the function are tending toward a specific value, c, and $x_n$ and $y_n$ $\to c$ shouldn't $f(z_n) \to L_1$ and $L_2$? $\endgroup$ – user23793 Nov 21 '12 at 18:52

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