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

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

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

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

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

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

So the epsilon-delta game begins with epsilon.

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

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

To put it all in one place you have:
For arbitrarily selected $\varepsilon>0$, you can use any $\delta \in \left\langle 0, -4 +4\sqrt{1+\frac{\varepsilon}{16}}\right\rangle$ and the inequality $|x-3|<\delta$ would imply that the inequality $|f(x)-f(3)|<\varepsilon$ 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 $\varepsilon=9$ then you should use $\delta \in \langle 0, -4+4\sqrt{1+\frac{9}{16}}\rangle=\langle 0,1\rangle.$

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

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

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

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

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