# weaker condition for continuity.

• May 23rd 2011, 10:55 AM
abhishekkgp
weaker condition for continuity.
Let $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}, \, a\in S \subset \mathbb{R}$.Suppose $\displaystyle f$ has the property that:
$\displaystyle x_n \in S, \, x_n \rightarrow a \Rightarrow (f(x_n))$ is convergent. Prove that $\displaystyle f$ is continuous at $\displaystyle a$.

my approach:
Let $\displaystyle (x_n)$ and $\displaystyle (y_n)$ be two sequences in $\displaystyle S$ with $\displaystyle x_n \rightarrow a, \, y_n \rightarrow a$. let $\displaystyle f(x_n) \rightarrow L_1$ and $\displaystyle f(y_n) \rightarrow L_2$. I need to prove that $\displaystyle L_1=L_2$. how do i go about it?
• May 23rd 2011, 12:43 PM
tonio
Quote:

Originally Posted by abhishekkgp
Let $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}, \, a\in S \subset \mathbb{R}$.Suppose $\displaystyle f$ has the property that:
$\displaystyle x_n \in S, \, x_n \rightarrow a \Rightarrow (f(x_n))$ is convergent. Prove that $\displaystyle f$ is continuous at $\displaystyle a$.

my approach:
Let $\displaystyle (x_n)$ and $\displaystyle (y_n)$ be two sequences in $\displaystyle S$ with $\displaystyle x_n \rightarrow a, \, y_n \rightarrow a$. let $\displaystyle f(x_n) \rightarrow L_1$ and $\displaystyle f(y_n) \rightarrow L_2$. I need to prove that $\displaystyle L_1=L_2$. how do i go about it?

I think this is false. Take for example $\displaystyle \displaystyle{f(x):=\left\{\begin{array}{cc}\frac{ \sin x}{x}&\mbox{ , if } x\neq 0\\{}\\ 8&\mbox { , if } x=0\end{array}\right.}$ , and let

$\displaystyle a\in S:=(-1,1)\subset \mathbb{R}$ .

Tonio
• May 23rd 2011, 12:44 PM
Opalg
Quote:

Originally Posted by abhishekkgp
Let $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}, \, a\in S \subset \mathbb{R}$.Suppose $\displaystyle f$ has the property that:
$\displaystyle x_n \in S, \, x_n \rightarrow a \Rightarrow (f(x_n))$ is convergent. Prove that $\displaystyle f$ is continuous at $\displaystyle a$.

my approach:
Let $\displaystyle (x_n)$ and $\displaystyle (y_n)$ be two sequences in $\displaystyle S$ with $\displaystyle x_n \rightarrow a, \, y_n \rightarrow a$. let $\displaystyle f(x_n) \rightarrow L_1$ and $\displaystyle f(y_n) \rightarrow L_2$. I need to prove that $\displaystyle L_1=L_2$. how do i go about it?

You can do this by considering the sequence $\displaystyle x_1,y_1,x_2,y_2,x_3,y_3,\ldots$.

Notice that it is not sufficient just to show that $\displaystyle L_1=L_2$. You need to show that this limit is equal to $\displaystyle f(a)$. You can do that by taking one of your sequences to be the constant sequence $\displaystyle x_n=a$ for all n.
• May 23rd 2011, 10:09 PM
abhishekkgp
Quote:

Originally Posted by tonio
I think this is false. Take for example $\displaystyle \displaystyle{f(x):=\left\{\begin{array}{cc}\frac{ \sin x}{x}&\mbox{ , if } x\neq 0\\{}\\ 8&\mbox { , if } x=0\end{array}\right.}$ , and let

$\displaystyle a\in S:=(-1,1)\subset \mathbb{R}$ .

Tonio

a typo error in my question.... i had written $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}$ while it should be $\displaystyle f:S \rightarrow \mathbb{R}$.