how to demonstrate that $\lim_{x\to2}3x^2=12$ using $\gamma$ and $\epsilon$ definition

Question

how to demonstrate that $\lim_{x\to2}3x^2=12$ using $\gamma$ and $\epsilon$ definition?

My current steps I have in order from top to bottom:

$|3x^2-12|<\epsilon$ and $0<|x-2|<\gamma$

assume $z=x-2$

$0<|z|<\gamma$ and $3z^2+12|z|<\epsilon$

$3z^2+12|z|<3\gamma^2+12\gamma=3\gamma(4+\gamma)$

Then assume $\gamma<1$

$\gamma=15$ hence $\gamma=\frac{\epsilon}{15}$

then now i am stuck and have no clue how to proceed

Answer 1

Okay, let's just say that everything you've done is scratch paper. You have reached the conclusion $\delta$ should be $ \frac {\epsilon}{15}$ and $\delta < 1$. So know we need to turn this into a proof.

For any $\epsilon > 0$, we can define $\delta = \min (1, \frac {\epsilon}{15})$.

Now for any $x$ so that $|x-2| < \delta$ (that is to say for any $x \in (2-\delta, 2+ \delta)$ then

$-\delta < x- 2< \delta$

$0 < 2 - \delta < x < 2 + \delta$

$(2-\delta)^2 < x^2 < (2+\delta)^2$

$4 - 4\delta + \delta^2 < x^2 <4\delta + \delta^2$.

$-4\delta + \delta^2 < x^2 - 4 < 4 \delta + \delta^2$.

Now as $\delta \le 1$ we know $\delta^2 \le \delta$ and so

$-5\delta < -4\delta + \delta^2 < x^2 - 4 < 4 \delta + \delta^2 < 5\delta$

$-15\delta < 3x^2 - 12 < 15\delta$

$|3x^2 -12| < 15\delta$.

So $\delta \le \frac {\epsilon}{15}$ then

$|3x^2 - 12| < 15\delta \le 15\frac {\epsilon}{15}=\epsilon$.

And that is the definition of $\lim_{x\to 2} 3x^2 = 2$.

(I do advise rereading the definition and understanding what it means:

(That is for ANY small distance $\epsilon > 0$ no matter how small, we can determine that there is some distance $\delta > 0$ so then whenever we can choose ANY $x$ within $\delta$ of $2$, that is to say, if $|x -2| < \delta$, we will then know that it MUST also be that $3x^2$ is within $\epsilon$ of $12$. That is $|3x^2 - 12| < \epsilon$.)

Answer 2

Let $|x-2|\lt 1$ , then

$-1<x-2<1$, or $3<x +2<5$.

$|3x^2-12| = 3|x-2||x+2|$.

Let $\epsilon >0$ be given.

Choose $ \delta = \min (1,\epsilon/(15))$.

Then $|x-2| \lt \delta$ implies

$3|x-2||x+2| \lt $

$(3)(5)|x-2| \lt (15)\delta \lt \epsilon$.

Answer 3

What you have given is sometimes called a "synthetic" proof- you start with what you want to prove, $|3x^2- 12|< \epsilon$, and arrive at $\gamma\le \frac{\epsilon}{15}$. Of course, you can't prove something by assuming it is true! This is, still, a valid proof because every step is reversible. That is, the "real proof" starts by taking $\gamma$ to be any number less than $\frac{\epsilon}{15}$ then works backwards to $|3x^2- 12|< \epsilon$.