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I am having trouble understanding part of the solution to this simple problem.

$\lim_{x \to 2} (x^2 + 3x) = 10$


Let $\epsilon > 0$

$| x - 2 | < \delta$ and $| x^2 +3x -10 | < \epsilon$

since $x^2 +3x -10 = (x - 2)^2 + 7x -14 = (x - 2)^2 + 7x -14 = ( x -2 )^2 +7(x-2)$

$|(x-2)^2 +7(x-2)| \leq |(x-2)|^2 +7|(x-2)|$

$\delta^2 + 7\delta < \epsilon$

let $\delta$ be the minimum of $1$ and $\epsilon/8$, $\delta^2 \leq \delta$.

then $8\delta < \epsilon$

$\delta < \epsilon/8$.

My Question:

I worked my way through the question down to $\delta^2 + 7\delta < \epsilon$

I then got confused by the end of this statement

Let $\delta$ be the minimum of $1$ and $\epsilon/8$, $\delta^2 \leq \delta$.

and in particular $\delta^2 \leq \delta$. I see how this allows me to prove the limit but I cannot make sense out of $\delta^2 \leq \delta$.

Could anyone explain this to me?

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You can use $\TeX$ on this site by enclosing formulas in dollar signs; single dollar signs for inline formulas and double dollar signs for displayed equations. You can see the source code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". Here's a basic tutorial and quick reference. There's an "edit" link under the question. – joriki Dec 23 '12 at 22:38
@joriki thanks.just seen tried to edit my question, but seen that you be me to it :-D – MWright Dec 23 '12 at 22:41
Also, for epsilon $\epsilon$, use \epsilon instead of $\in$: \in, which denotes "is and element of". – amWhy Dec 23 '12 at 22:44
@amWhy thanks will edit – MWright Dec 23 '12 at 22:46
Well $0 < \delta \le 1$ so by multiplying by $\delta$ we get $0 < \delta ^ 2 \le \delta$ – xavierm02 Dec 23 '12 at 22:48
up vote 3 down vote accepted

By choosing $\delta$ the minimum of $1,\frac{\epsilon}{8}$ no matter what the value of $\epsilon$:

$\delta \leq1 \Rightarrow \delta^2 \leq \delta \Rightarrow \delta^2+7\delta\leq 8\delta$

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We have been challenged with an $\epsilon$, perhaps $\epsilon=1/1000$. We want to come up with a $\delta$ such that if $|x-2|\lt \delta$, then for sure $|x^2+3x-10|\lt \epsilon$.

Suppose that after some calculation, we announce that $\delta=\epsilon/8$ does the job. Then triumphantly the challenger could say that she had $\epsilon=47$ in mind, and that in that case taking $\delta=47/8$ is insufficient. Of course, that is not playing fair. But we might as well come up with a $\delta=\delta(\epsilon)$ that always works.

Some algebra shows that $x^2+3x-10=(x-2)^2+7(x-2)$. So we want to make sure that $|(x-2)^2+7(x-2)|\lt \epsilon$. Note that $$|(x-2)^2+7(x-2)|\le (x-2)^2+7|x-2|.$$ We want to make the right-hand side "small," by choosing $x$ appropriately close to $2$. Suppose we had been given a ridiculous $\epsilon$, like $47$. If $x$ is within $47$ of $2$, the number $(x-2)^2$ could be very large. So the first task is to make sure $\delta$ is small enough not to allow $(x-2)^2$ to be large.

So we say first of all, let $\delta\le 1$. Then if $|x-2|\lt \delta$, it follows that $(x-2)^2\lt \delta$. For $(x-2)^2=(x-2)(x-2)$. The "first" $x-2$ has absolute value $\le 1$, and the second has absolute value $|x-2|$, so the product has absolute value $\le |x-2|$.

It follows that as long as $\delta\le 1$, we have $(x-2)^2+7|x-2|\le 8|x-2|$. To make sure this is $\lt \epsilon$, it is enough to make $\delta=\frac{1}{8}\epsilon$.

However, in deriving our simplified inequality, we assumed that $\delta\le 1$. So we know that everything will work if $\delta=\min(1,\epsilon/8)$.

It might have been better to observe that $$(x-2)^2+7|x-2|=|x-2|\left(|x-2|+7\right).$$ Then it is clear that if $|x-2|\le 1$, then $(x-2)^2+7|x-2|\le 8|x-2|$.

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