Prove that $\lim_{x \to 2} 5x^2 = 20$ using $\epsilon - \delta$ definition. 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 $\delta$.

 A: 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
$$\delta=\min(\frac 13,\frac{\epsilon}{5(\frac 13+4)})$$
A: The author is looking for a $\delta > 0$ for which $|x - 2| < \delta \Rightarrow |x^2 - 25| < \epsilon$. If some $\delta$ works, then a smaller $\delta$ works as well and by assuming $|x - 2| < 1$ the author is looking for a $\delta$ that is at most $1$.
That is what he is saying with "Take $\delta$ to be smaller than both $1$ and $\epsilon/25$".
To explicitly answer your first question: the author claims "if $|x - 2| \leq 1$, then $|x + 2| \leq 5$ and ... $|f(x) - L| \leq 25|x-2|$", so the assumption $|x - 2| \leq 1$ is discharged right there; in the next sentence that is already not assumed anymore.
