Proof of Epsilon - Delta Limit

Question

I am trying to prove the limit of the following:

$\lim x \rightarrow 1 $ of $x^2+2x$ where the limit $L = 3$

This gives us $0<|x-1|<\delta$ and $|x^2+2x-3|<\epsilon$

First I factorize $f(x) \rightarrow$ $|(x+3)(x-1)|<\epsilon$

$\rightarrow |x+3||x-1|<\epsilon $

I recognize that I get a matching term $|x-1|$, however I have an uncontrolled term $|x+3|$.

$|x-1|<\epsilon/|x+3| $

If I assume $|x-1|<\delta$ and $\delta < 1$

Edit, made some progress;

I.e. making a restriction that $x$ can be a maximum distance of 1 away from $a$.

$|x-1| < 1$

$0< x<2$

$3<x+3<5$

Thus, $\epsilon/|x+3|$ is at its minimum when $|x+3|$ is at its maximum

$|x-1|< \frac\epsilon5$

Answer 1

Hint: You can add a condition depending on the size of $\epsilon$. i.e change your $\delta$ depending on $\epsilon$. What happens if you let

$$\delta = \min\{1,\frac{\epsilon}{5}\}?$$

If $|x-1| < \delta$ then what do you know about $|x+3|$ and $|x-1|$ in terms of constants and $\epsilon$?

Edit: Here is a full solution at the OPs request.

Let $f(x) = x^2 +2x$. Given $\epsilon > 0$ define $\delta = \min\{1,\frac{\epsilon}{5}\}$. We note that

\begin{align*} |f(x) -1| &= |x^2 + 2x -3|\\ &=|x+3||x-1|. \end{align*} When $|x-1| < \delta$ we have $|x-1| < \frac{\epsilon}{5}$ and $|x+3| <5$ and hence

\begin{align*} |f(x)-1|&=|x+3||x-1|\\ &\le 5 \cdot \frac{\epsilon}{5}\\ &=\epsilon \end{align*}

We have shown that for any $\epsilon > 0$ there exists $\delta = \min\{1,\frac{\epsilon}{5}\}$ such that when $|x-1|<\delta$, $|f(x) -1| < \epsilon$. This implies that $f(x) \rightarrow 1$ when $x \rightarrow 1$.

Answer 2

Continuing from my original post

$|x-1|<\epsilon/|x+3|<\frac\epsilon5$

We have two restrictions:

1) $|x-1|<1$

2) $|x-1|<\frac\epsilon5$

Let $\delta=$ min{$1,\frac\epsilon5$}

Proof:

Given a $\epsilon>0$, and let $\delta =\frac\epsilon5$

We know that $3<x+3<5$, thus:

$|x-1|<\frac\epsilon5$

$\rightarrow |x-1|(x+3)<\frac\epsilon5(5)$ (*)

$\rightarrow |x^2+2x-3|<\epsilon$

$\rightarrow |f(x)-L|<\epsilon$

(*) This part makes sense, however I am not 100 % sure why we do it. I just got it from another example. Why do we multiply by x+3 and 5 (different values)? I know that $x+3 < 5$ but still...

Now let $\delta=1$ (BTW is it common to prove for both delta values?

$|x-1|<1$

$|x-1|(x+3)<(x+3)$ here we multiply by the same quantity for some reason...

$|x^2+2x-3|<x+3<5<\epsilon$

$|x^2+2x-3|<\epsilon$

And I think that's it. This is my first encounter with proofs actually, so I'm still a noob. All tips are welcome.