1. ## Continuity proof

Show that f(x) = x^2 + 2x is continuous at 3.

I know that I need to show that if |x - 3| < \delta then |x^2 +2x -15| < \epsilon. But I am not sure what to do after this.

2. $\displaystyle |x^2+2x-15|=|(x-3)(x+5)|=|x-3||x+5| < \epsilon$

If we demand that $\displaystyle |x-3|< 1$, i.e. $\displaystyle 2<x<4$, then $\displaystyle |x+5|<9$.

If we also demand that $\displaystyle |x-3|<\frac{\epsilon}{9}$, then we get

$\displaystyle |(x-3)(x+5)|=|x-3||x+5| < 9\cdot \frac{\epsilon}{9}=\epsilon$

If we let $\displaystyle \delta$ be the smallest of the two numbers $\displaystyle 1$ and $\displaystyle \frac{\epsilon}{9}$, then we have $\displaystyle |x^2+2x-15|=|(x-3)(x+5)| < \epsilon$ when $\displaystyle 0<|x-3|<\delta$

3. To be precise it has to be proven that for any real number $\displaystyle \varepsilon>0$, a real number $\displaystyle \delta >0$ can be found such that the implication $\displaystyle |x-3|<\delta \Rightarrow |f(x)-f(3)|<\varepsilon$ holds.

So the epsilon-delta game begins with epsilon.

$\displaystyle |f(x)-f(3)|<\varepsilon \Rightarrow |x-3||x+5|<\varepsilon,$
$\displaystyle |x-3||x-5|=|x-3||x-3+8|\leq |x-3|(|x-3|+|8|)=|x-3|^2+8|x-3|$

Now, for arbitrarily chosen $\displaystyle \varepsilon$ look at the inequality $\displaystyle 0<|x-3|^2+8|x-3|<\varepsilon$. Treat the expression $\displaystyle |x-3|$ as you would any other normal variable (think of it as $\displaystyle \delta$) and you have a quadratic inequality which holds when $\displaystyle |x-3|\in \left\langle 0, -4 +4\sqrt{1+\frac{\varepsilon}{16}}\right\rangle$.

To put it all in one place you have:
For arbitrarily selected $\displaystyle \varepsilon>0$, you can use any $\displaystyle \delta \in \left\langle 0, -4 +4\sqrt{1+\frac{\varepsilon}{16}}\right\rangle$ and the inequality $\displaystyle |x-3|<\delta$ would imply that the inequality $\displaystyle |f(x)-f(3)|<\varepsilon$ holds. Thus you have proven the continuity at 3.

What Mondreus used as an example is simply a special case of this story: were you to choose $\displaystyle \varepsilon=9$ then you should use $\displaystyle \delta \in \langle 0, -4+4\sqrt{1+\frac{9}{16}}\rangle=\langle 0,1\rangle.$

4. This makes sense to me, but I am not sure how you come up with |x - 3| < 1. Why 1? Thanks.

Originally Posted by Mondreus
$\displaystyle |x^2+2x-15|=|(x-3)(x+5)|=|x-3||x+5| < \epsilon$

If we demand that $\displaystyle |x-3|< 1$, i.e. $\displaystyle 2<x<4$, then $\displaystyle |x+5|<9$.

If we also demand that $\displaystyle |x-3|<\frac{\epsilon}{9}$, then we get

$\displaystyle |(x-3)(x+5)|=|x-3||x+5| < 9\cdot \frac{\epsilon}{9}=\epsilon$

If we let $\displaystyle \delta$ be the smallest of the two numbers $\displaystyle 1$ and $\displaystyle \frac{\epsilon}{9}$, then we have $\displaystyle |x^2+2x-15|=|(x-3)(x+5)| < \epsilon$ when $\displaystyle 0<|x-3|<\delta$