Here is a proof for $\lim_{x\to2} 3x^2 = 12$
We are given some $\epsilon > 0$, and we need to find $\delta$ such that
$0 < |x-2| < \delta \Rightarrow |3x^2 – 12| < \epsilon$
The inequality $|3x^2 – 12| < \epsilon$ will be more useful if it is in terms of $x-2$ rather than x, since the inequality $0 < |x-2| < \delta$ is in terms of $x-2$. For simplicity, let $z = x-2$. Then we wish to find $\delta$ such that
$0 < |z| < \delta \Rightarrow |3(z+2)^2 – 12| < \epsilon$
We can simplify this to
$0 < |z| < \delta \Rightarrow |3z^2 + 12z| < \epsilon$
However, we know that $|3z^2 +12z|\leq|3z^2|+|12z|=3z^2 + 12|z|$. So it suffices to find $\delta$ such that
$0 < |z| < \delta \Rightarrow 3z^2 + 12|z| < \epsilon$
If $ 0 < |z| < \delta$, then $3z^2 + 12|z| < 3\delta^2 + 12\delta = 3\delta(4 + \delta)$. Thus it suffices to choose $\delta$ such that
$3\delta(4 + \delta) < \epsilon$
The $4 + \epsilon$ term is somewhat annoying. We can make it simpler by assuming that $\delta \leq 1$.
If we assume that $\delta \leq 1$, then $4+\delta \leq 5$, and the inequality that we need becomes simply
$3\delta(4+\delta) \leq 15\delta < \epsilon$
To force this to be true, we select $\delta = \frac{\epsilon}{15}$. (In the unlikely event that $\epsilon > 15$, we can just take $\delta = 1$.) We then conclude that
$0 < |z| < \delta \Rightarrow |3x^2 – 12| < 3\delta(4 + \delta) \leq 15\delta=\epsilon$.
Thus, for any $\epsilon < 15$, we have found that $\delta = \frac{\epsilon}{15}$ satisfies the $\delta$-$\epsilon$ condition:
$0 < |x-2| < \delta \Rightarrow |3x^2 – 12| < \epsilon$
and hence we have stablished that $\lim_{x\to2} 3x^2 = 12$
What I don't understand is why just because something is greater than $|3z^2 + 12z|$ It suffices to find $\delta$ using it. I mean by that logic couldn't I say something like
$|3z^2 + 12z| \leq |3z^2 + 12z| + z^{5000000}$, so it suffices to find $\delta$ such that $0 < |z| < \delta \Rightarrow|3z^2 + 12z| + z^{5000000} < \epsilon$
Any help would be appreciated
Best Answer
[Not a direct answer to your question. But too long to be a comment.] I appreciate your effort writing down the long question. But I strongly dislike the way your textbook gives the proof: it makes things look so complicated and that's not the way we do analysis in practice!
Here is what one could do. Let $0<\epsilon<1$. One wants to find $\delta>0$ so that the following implication is true $$ 0<|x-2|<\delta\Rightarrow |x^2-4|<\epsilon. $$ Observe that $|x^2-4|<\epsilon$ is equivalent to $$ |x-2|\cdot|x+2|<\epsilon\tag{1} $$ One can see that when $x$ is getting "close" to $2$, $|x-2|$ can be as small as possible while $|x+2|$ remains bounded by some fixed number. This is the essential point to give the proof. To make it precise, choose $\delta=\epsilon$. Then if $0<|x-2|<\delta$, we have $$ |x-2|\cdot |x+2|\leqslant\epsilon (\epsilon +4)<5\epsilon\tag{2} $$ where we use the triangle inequality $|x+2|\leqslant |x-2|+4$. I claim that (2) completes the proof by the following easy exercise.
I would like to repeat a remark I made in another answer:
One important tactic that is seldom mentioned in elementary real analysis or calculus textbooks is that when doing an estimate in analysis, one should never worry about the constant in front of one's epsilon. As Terry Tao points out in one of his excellent blog posts on problem solving strategies in real analysis: