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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

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

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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.

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  • $\begingroup$ Great to see you answer your own question! +1 $\endgroup$ Sep 5 '16 at 14:58
  • $\begingroup$ The way you've written your answer is confusing - why have you written $\delta = \frac{\epsilon}{5}$ in the line under 'Proof'? The line after this one (the one about $x+3$ being bounded) does not follow with such $\delta$. $\endgroup$ Sep 5 '16 at 14:59
  • $\begingroup$ I might be mistaken, but as I understand it, you are supposed to prove for different values of $\delta$? Let me re-write it; I added a "let", does that make more sense? $\endgroup$
    – themli
    Sep 5 '16 at 15:06
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    $\begingroup$ The idea is to find one $\delta$ that works for a given $\epsilon$. What my answer is trying to say, is that your $\delta$ may be different for different $\epsilon$. Does this make sense? $\endgroup$ Sep 5 '16 at 15:14
  • $\begingroup$ Yes it makes sense. The $\delta-\epsilon$ definition shows that for any given $\delta$, I can provide you with a matching $\epsilon$, right? But I don't understand what I would change about my proof. Edit: So you are saying I should eliminate the part where delta = 1? Since delta should be one specific value? $\endgroup$
    – themli
    Sep 5 '16 at 15:32
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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 $x \rightarrow 1$.

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  • $\begingroup$ Well then; $0<|x-1|<\epsilon/5$ and $|x+3| |x-1| < \epsilon$ which means that it is greater by a factor of 5. Thus, $5|x-1|<\epsilon$ $\rightarrow |x+3||x-1|<\epsilon$ $\rightarrow |x+3|<5$ $\rightarrow x<2$ Not sure if I'm getting anywhere, but I'll read through the PDF posted by @NickLiu $\endgroup$
    – themli
    Sep 5 '16 at 13:29
  • $\begingroup$ Good thinking - $|x-1| < \frac{\epsilon}{5}$ if $|x-1|< \delta$. Notice also that $|x+3| <5$ when $|x-1|< \delta$. Overall this implies $|x-1||x+3|<5 \cdot \frac{\epsilon}{5} = \epsilon$ when $|x-1|<\delta$. $\endgroup$ Sep 5 '16 at 14:46

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