Prove rigorously: $\displaystyle \lim_{x\rightarrow 2}x^2+5x=14$

Question

Prove rigorously: $\displaystyle \lim_{x\rightarrow 2}x^2+5x=14$

I am using an epsilon-delta proof, and this is as far as I got:

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

$|x^2+5x-14|<\epsilon$

$|x-2||x+7|<\epsilon$

And I bound $|x+7|$,

$0<|x-c|<\delta$

$0<x-2<1$

Add $9$ to all sides.

$10<x+7<11$

And, according to my instructor, I assume the largest number because it is not in the denominator,

$|x-2||x+7|<\epsilon$

$|x-2|\cdot11<\epsilon$

$|x-2|<\frac{\epsilon}{11}$

How do I finish the proof?

Answer 1

We want that given $\epsilon$, then $|x-2|<\delta $ imply $|x^2+5x-14|<\epsilon$, then $x-2<\delta$ adding $9$, $x+7<\delta +9$ then $|x-2||x+7|<(9+ \delta)\delta $, then you should take $\epsilon = (9+\delta )\delta$. So you take the positive root of the polynomial equation to find $\delta $ approprieted. It's possible, because you need to think only on little value for $\epsilon$.

Answer 2

The crux of these types of proofs is to demonstrate how we can always choose a $\delta$ as a function of $\epsilon$. We call these kinds of functions Skolem Functions.

Lets look at the full definition of the limit for your problem:

$\forall \epsilon >0 \exists \delta > 0 \forall x : (|x-2|<\delta) \rightarrow (|x^2 +5x - 14| < \epsilon)$

What we need to demonstrate is the $f(\epsilon)$ that satisfies

$\forall \epsilon >0 \forall x : (|x-2| < f(\epsilon)) \rightarrow (|x^2 + 5x -14| < \epsilon)$

As André Nicolas pointed out, the function $f(\epsilon) = min(1,\epsilon/10)$ works.

Why? How does this function relate to the derivative of your function about the limit point? What value(s) of $n$ could you use if your function was $f(\epsilon) = min(2,\epsilon/n)$ ?

Answer 3

The following answer is based on the solution by math_man:

Given $\epsilon>0$, choose any $\delta$ satisfying $\displaystyle 0<\delta<\frac{\sqrt{81+4\epsilon}-9}{2}$.

Then $2\delta+9<\sqrt{81+4\epsilon}\implies 4\delta^{2}+36\delta+81<81+4\epsilon\implies\delta^2+9\delta<\epsilon$, so

$0<|x-2|<\delta\implies \lvert x^2+5x-14\rvert=\lvert x-2\rvert\lvert x+7\rvert\lvert<\delta\lvert(x-2)+9\rvert\le\delta(\lvert x-2\rvert+9)<\delta(\delta+9)<\epsilon$