# Thread: Prove Function is Continuous using epsilon/delta

1. ## Prove Function is Continuous using epsilon/delta

Here is the theorem i'm supposed to use to prove the continuity of the function f(x) = x^3.

Let f be a real-valued function whose domain is a subset of R. Then f is continuous at x0 element dom(f) iff for every epsilon > 0 there exists delta > 0 s. t. x element dom(f) and |x - x0| < delta => |f(x)-f(x0)| < epsilon.

So I don't know how to get delta. I'm pretty sure you are supposed to use the triangle inequality somewhere, and you need to work backwards but i'm just a bit confused.
Thanks

2. Originally Posted by tbyou87
Here is the theorem i'm supposed to use to prove the continuity of the function f(x) = x^3.

Let f be a real-valued function whose domain is a subset of R. Then f is continuous at x0 element dom(f) iff for every epsilon > 0 there exists delta > 0 s. t. x element dom(f) and |x - x0| < delta => |f(x)-f(x0)| < epsilon.

So I don't know how to get delta. I'm pretty sure you are supposed to use the triangle inequality somewhere, and you need to work backwards but i'm just a bit confused.
Thanks
I will do the hard part.

Let $\displaystyle \epsilon > 0$. Choose some $\displaystyle \delta > 0$ to be determined, such that $\displaystyle |x - x_0| < \delta$ when $\displaystyle x \in dom(f)$. Furthermore, we may need an upper bound for $\displaystyle |x + x_0|$ given that $\displaystyle |x - x_0|< \delta$ so let us say $\displaystyle |x - x_0|<1 \implies |x|<|x_0| + 1$. So , by the triangle inequality, $\displaystyle |x + x_0| \le |x| + |x_0| < 2 |x_0|+1$. Now consider $\displaystyle |f(x) - f(x_0)|$. We have:

$\displaystyle |x^3 - x_0^3| = \left|(x - x_0) \left( x^2 + xx_0 + x_0^2 \right) \right| \le$ $\displaystyle \left|(x - x_0) \left( x^2 + 2xx_0 + x^2\right) \right| = |(x - x_0)(x + x_0)^2| < |x - x_0|(2|x_0| + 1)^2$

Now, what does $\displaystyle \delta$ have to be to make that last part
$\displaystyle < \epsilon$?

3. Ok so I think delta = min{1, epsilon/(2|x0|+1)^2}

I still have a question about why you thought of using the upperbound for |x+x0| and why you assume delta=1 at the beginning and not another value?
Also, is that the general formula for solving these continuity problems?
Thanks

4. Originally Posted by tbyou87
Ok so I think delta = min{1, epsilon/(2|x0|+1)^2}
correct

I still have a question about why you thought of using the upperbound for |x+x0|
i cheated, well not really, it's like when you want to find delta, you work backwards, so too did i work backwards here. i simplified |f(x) - f(x_0)| and realized i would need a bound for that

and why you assume delta=1 at the beginning and not another value?
because i felt like it. choosing 1/2 would work as well, so would 1/3, 2/5 ...

1 is just a nice number to use, and it's the number my professor always uses

Also, is that the general formula for solving these continuity problems?
Thanks
yes, this is more or less the general approach to this problems. there are both simpler and harder types to prove where you would have to modify your technique somewhat

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