Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm self-studying from the book Understanding Analysis by Stephen Abbott, and I don't understand example 4.2.2 on page 105.

The aim of the example is to show that: $$ \lim_{x \to 2} g(x) = 4 $$ where $g(x) = x^2$.

The author starts by writing: $$ |g(x) - 4| = |x^2 - 4 | = |x+2||x-2| $$ which I understand. But then the following paragraph makes no sense to me:

We can make $|x-2|$ as small as we like, but we need an upper bound on $|x+2|$ in order to know how small to choose $\delta$. The presence of the variable $x$ causes some initial confusion, but keep in mind that we are discussing the limit as $x$ approaches $2$. If we agree that our $\delta$-neighborhood around $c = 2$ must have radius no bigger than $\delta = 1$, then we get the upper bound $|x + 2| \leq |3 + 2| = 5$ for all $x \in V_\delta(c)$.

By saying we are only considering the $\delta$-neighborhood around $c=2$ with a radius smaller than or equal to $1$, I would think we get: $$ V_\delta (c) = \{ x \in \mathbb{R} \mid |x-2| < 1 \} = (1,3) $$ so I don't understand where "$|x + 2| \leq |3 + 2| = 5$ for all $x \in V_\delta(c)$" comes from? Any help is much appreciated.

share|improve this question
3  
If $\lvert x-2\rvert < 1$, that is, $1 < x < 3$, then $3 < x+2 < 5$, hence $\lvert x+2\rvert \leqslant 5$. –  Daniel Fischer Aug 29 at 20:58
    
The first inequality is $1\lt x\lt 3$. @DanielFischer –  Fermat Aug 29 at 20:59
    
@DanielFischer thanks! –  Hunter Aug 29 at 21:10

2 Answers 2

up vote 3 down vote accepted

I like using the triangle inequality here instead: $$ |x + 2| = |(x - 2) + (4)| \leq |x - 2| + |4| < 1 + |4| = 5 $$

share|improve this answer
    
This really clarifies it and is very clean. Thanks! –  Hunter Aug 29 at 21:07

if $x \in V_{\delta}(c)$, we have $|x-2| < 1$. Since $||x|-|2|| \leq |x-2|$ we deduce: $|x|-2 \leq ||x|-|2|| \leq |x-2| < 1$. Then $|x| < 1+2=3$.

Now : $$|x+2| \leq |x| + |2| < 3+2=5$$

share|improve this answer
    
Thanks! I've accepted Adriano's answer because it cleaner, but this is still very useful. +1 –  Hunter Aug 29 at 21:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.