Proof of $\lim_{x\to 4}x^2 = 16$ using the definition of a limit

Question

Hi I watched this video, and at 11:32, the professor concluded that $9|x-4| < ε$. The way I see it: $|x+4| < 9 ⟺ |x-4||x+4| < 9|x-4| \;∀ x ≠ 4\;$ (1). We also have $|x-4||x+4| < ε \;$ (2). From (1) and (2), it doesn't seem enough to conclude that $9|x-4| < ε$. Am I missing something?

I was able to come up with a proof, but it doesn't look very neat, I wonder if there may be a better way, but here is mine:

Set δ = $\sqrt {ε +16} - 4\qquad (ε > 0) $

When $0 < |x-4| < δ$,

we have: $0 < |x-4| < \sqrt {ε +16} - 4$

$⟺ -(\sqrt {ε +16} -4) < x - 4 < \sqrt {ε +16} -4\qquad (x ≠ 4)$

$⟺ 8 - \sqrt {ε +16} < x < \sqrt {ε +16}\qquad(x ≠ 4)$

$⟺ - \sqrt {ε +16} < 8 - \sqrt {ε +16} < x < \sqrt {ε +16}\qquad(x ≠ 4)$

$⟹ |x| < \sqrt {ε +16}\qquad(x ≠ 4)$

$⟺ -16 ≤ x^2 -16 < ε\qquad(x ≠ 4)$ (1)

------(Now I'm going to prove $- ε < x^2 - 16$)------

• When $0 < ε ≤ 48,\quad 0 ≤ 8 - \sqrt {ε +16} < x$

$⟺ (8 - \sqrt {ε +16})^2 - 16 < x^2 - 16$ (2)

Consider $f(ε) = [(8 - \sqrt {ε +16})^2 - 16] -(-ε)$

$f(ε) = 2(\sqrt {ε +16} - 4)^2 > 0 $

$⟹ -ε < (8 - \sqrt {ε +16})^2 - 16$ (3)

(2),(3) $⟹ -ε < x^2 - 16$ (4)

(1),(4) $⟹ -ε < x^2 - 16 < ε ⟺ |x^2 - 16| < ε$ (i)

• When $ε > 48,\quad -ε<-16 ≤ x^2 -16 < ε ⟺ |x^2 - 16| < ε$ (ii)

(i),(ii) $⟹ |x^2 - 16| < ε \quad ∀ x\;:\; 0 < |x-4| < δ,\;(δ = \sqrt {ε +16} - 4)$

Therefore, $\lim_{x\to 4}x^2 = 16$

Please let me know if there is a better way. Thank you!

Answer 1

$|x-4||x+4|\leq |x-4|(|x-4|+8)<|x-4|(1+8)$ if $|x-4|<1$. So $|x-4||x+4|<\epsilon$ if $|x-4|<\frac {\epsilon } 9$ and $|x-4|<1$. Take $\delta =\min \{1, \frac {\epsilon } 9\}$.

Answer 2

The video proof explicitly states that if $\delta < 1$, then $|x+4|<9$.

But remember, that is only the "scratch" part of the video. It is the part where we are looking for what we should set $\delta$ to. Once we decide on a delta, we need to also prove that it is correct.

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As for your proof, it looks ok but it is quite messy. I think the video proof is much nicer. Remember, the proof part of the video is very short (once you come up with the delta), the main part is just looking for $\delta$.

The proof is quite simple:

1. Let $\epsilon > 0$.
2. Let $\delta = \min\{1, \frac\epsilon9\}$.
3. Let $x$ be any value such that $0<|x-4|<\delta$.
4. From (3) we can conclude that $|x+4|<9$.
5. Then (I am writing the reasoning behind each (in)equality alongside it): $$\begin{align}|f(x)-16| &= |x^2-16| \\ &= |x-4||x+4|&\text{algebraic manipulation}\\ &\leq 9|x-4|&\text{from (4)}\\ &\leq 9 \delta&\text{from (3)}\\ &\leq 9\frac\epsilon9&\text{from (2)}\\ &=\epsilon\end{align}$$
6. Thus, $|f(x)-16|<\epsilon$, which is what we wanted to show.