Working on the book: Richard Hammack. "Book of Proof" (p. 259)
Example 13.2 Prove that $\lim_{x \to 2} 5x^2 = 20$.
Proof. Suppose $\epsilon$ > 0. Notice that
$$
| f(x) – L| = |5x^2 – 20| = |5(x^2 – 4)| = |5(x – 2)(x + 2)| = 5 · |x-2| · |x + 2|.
$$
Now we have a factor of $|x-2|$ in $|f(x)-L|$, but it is accompanied with $|x+2|$. But if $|x-2|$ is small, then $x$ is close to 2, so $|x+2|$ should be close to 4.
Now, the author assumes $|x-2| \leq 1$
In fact, if $|x-2| \leq 1$, then |$x+2| = |(x-2)+4| \leq |x-2|+|4| \leq 1+4 = 5$. (Here we applied the inequality (13.2) from page 245.) In other words, if $|x – 2| \leq 1, \text{then } |x + 2| \leq 5$, and the above equation yields
$$
| f (x) – L| = |5x^2 – 20| = 5 · |x – 2| · |x + 2| < 5 · |x – 2| · 5 = 25|x – 2|.
$$
Take $\delta$ to be smaller than both 1 and $\frac{\epsilon}{25}$ . Then $0<|x-2|<\delta$ implies $|5x^2-20|<25·|x-2|<25\delta<25\frac{\epsilon}{25}=\epsilon$. By Definition 13.2, $\lim_{x \to 2} 5x^2 = 20$
My questions are:
- Where does the assumption $|x-2| \leq 1$ come from and how it gets discharged ?
- I wonder if the author is really proving $\forall \epsilon > 0 ( \exists \delta > 0(|x-c| < \delta \Rightarrow (|x-c| \leq 1 \Rightarrow |f(x) – f(c)| < \epsilon)))$. Perhaps, I am wrong but it is not possible to add that assumption as a premise, as arbitrary variable $x$ appears after the introduction of $\delta$.
Best Answer
As we look for the limit when $ x $ goes to $ x=2 $, we can assume that $ x $ is not far from $ 2 $, let say $$|x-2|<a \; or\; 2-a<x<2+a\; $$ $$or\; -4-a<4-a<x+2<4+a$$
$$or \; |x+2|<4+a$$
with $ a>0$.
with this additional condition, given $\epsilon>0$, we look for $ \delta>0$ such that
$$|x-2|<a \; and\; |x-2|<\delta \implies$$ $$5 |x-2||x+2|<\epsilon$$
but $$5|x-2||x+2|<5(a+4)|x-2|$$
So, we just need to satisfy the condition $$|x-2|<\frac{\epsilon}{5(a+4)}$$
Thus, you just need that $$\delta=\min(a,\frac{\epsilon}{5(a+4)}).$$
The author prefered to choose $ a=1$
If you decide to take $ a=\frac 13$, you will choose $$\delta=\min(\frac 13,\frac{\epsilon}{5(\frac 13+4)})$$