Proving $\lim_{x \rightarrow 3} (x^3-9x)/(x-3)=18$ by definition

Question

(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.

Answer 1

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.

Answer 2

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!

Answer 3

HINT:

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