- Thread starter
- #1

Using an epsilon-delta technique, prove that f(x) = x

^{3}

is continuous at x = 1

Can someone provide a brief proof of this explaining the steps?

- Thread starter Tompo
- Start date

- Thread starter
- #1

Using an epsilon-delta technique, prove that f(x) = x

is continuous at x = 1

Can someone provide a brief proof of this explaining the steps?

- Feb 5, 2012

- 1,621

Hi Tompo,

Using an epsilon-delta technique, prove that f(x) = x^{3}

is continuous at x = 1

Can someone provide a brief proof of this explaining the steps?

Welcome to MHB! First you have to be familiar with the epsilon delta definition of continuity (Refer >>this<<).

We say that the function \(f:I\rightarrow \Re\) is continuous at \(c\in I\) if for each \(\epsilon>0\) there exists \(\delta>0\) such that,

\[| f(x) - f(c) |<\epsilon\mbox{ whenever }| x - c |<\delta\]

In your case you have \(f(x)=x^3\) and \(c=1\). First take any \(\epsilon>0\) and consider \(|f(x)-f(1)|\). Try to find a \(\delta>0\) such that \(|f(x)-f(1)|<\epsilon\) whenever \(|x-1|<\delta\). >>Here<< you will find some examples of using the epsilon delta definition to show continuity.

Kind Regards,

Sudharaka.

- Admin
- #3

- Jan 26, 2012

- 4,197

You can also check out the Differential Calculus Tutorial, in post # 2.