Jtdylktuy

(a) Find the following limit value, L: $$lim_{x rightarrow 3} (x^3-9x)/(x-3) = L$$
(b) Prove your result obtained in (a) above, using the precise definition of a limit (via the $varepsilon-delta$ formalism)

Solutions

I was able to find (a) $L = 18$, and I used the description in my notes to determine the proof, but I'm not quite sure of my working. Please view image for solution.

(b) $|f(x) - L| = |(x^3 - 9x)/(x-3) - 18| = |(x^3 - 27x + 54)/(x-3)| = |(x-3)(x+6)|$

I don't know where to go from here.

Since you want the find the limit at $x=3$, you have to prove that for $|x-3|<delta$ there exists an $epsilon >0$ such that $|f(x) - 18| < epsilon$.

Since $|f(x) - 18| = |(x-3)(x+6)|$, note that $|x-3| < delta$ by the initial hypothesis of the $epsilon-delta$ definition. Thus :

$$|f(x)-18| < delta|x+6|$$

But :

$$|x-3| < delta Leftrightarrow -delta < x-3 < delta Leftrightarrow 9-delta < x+6 < 9+delta $$

Thus

$$|f(x)-18|<delta(9+delta) equiv epsilon$$

since it can be made arbitrary small with $delta >0$ and the proof is completed.

For (a) notice that:
$$lim_{x rightarrow 3} frac{x^3 -9x}{x-3}=lim_{x rightarrow 3} frac{x(x+3)(x-3)}{x-3}= lim_{x rightarrow 3}x^2 +3x= 9+9=18. $$
So $L=18$, As you indeed got.

As for $(b)$ we often write out the definition of the limit if we want to do limit proofs, let $varepsilon$ be arbitrary, then for some $delta$, whenever $|x-3|< delta$, we need to show that $|f(x)-18|< varepsilon$, this part is very important we notice that this is the part that appears in your proof!, so we know that
$$|x-3| |x+6|< delta |x+6|$$

If we could only find some sort of bound on $|x+6|$, we could complete the proof and we would know what $varepsilon$ to pick. We notice that since $|x-3| < delta$ we can also say that $x in (3- delta, 3+ delta)$ now choose as a test $delta =1$ then we have that $x in (2, 4)$, in this way we can now say that:

$$ delta |x+6| <delta |4+6|=10 delta$$
Now if we pick $delta = frac{varepsilon}{10}$ we have a complete proof.
So let $delta =min(1, frac{varepsilon}{10})$ and you have a proof.

Let us complete your proof:
Let $epsilon >0$ be given, then let $delta = min(1, frac{epsilon}{10})$, notice that whenever $ |x-3|< delta $ we have that:
$$|f(x) - L| = left| frac{x(x-3)(x+3)}{x-3} -18 right| $$ $$= |x(x+3) - 18| = |x^2+3x-18| = |(x-3)(x+6)|< delta cdot 10 leq epsilon$$

Basically, the game is to estimate and bound $x$ in such a way that we can get a very nice expression for our absolute value expression and we get a nice relation of some number times $delta$. We then choose our $delta$ to be some function of $epsilon$ and we are done. That's the ultimate goal, to estimate and approximate until we are close enough. How close is close enough - well that's where a lot of the fun lies!

HINT:

Use that $$x^3+9x=x(x+3)(x-3)$$