# Find a δ δ such that | f ( x ) − L | < ϵ |f(x)−L|<ϵ for all x satisfying 0 < | x − a | < δ 0<|x−a

1. Aug 18, 2016

### Aliax3012

I've pondered at this problem for a long time, and I don't know where I make a mistake, can anyone give me a hint?

1. The problem statement, all variables and given/known data

find a $\delta$ such that $|f(x)-L|<\epsilon$ for all x satisfying $0<|x-a|<\delta$

2. Relevant equations

$f(x)=x^4$; $l=a^4$

3. The attempt at a solution

writing as the definition of limits

if $|x-a|<\delta$, then $|x^4-a^4|<\epsilon$

factor the right side of equation I got

$|(x^2-a^2)(x^2+a^2)|<\epsilon$, then I can have $|x^2-a^2|<\delta_1$

chose $\delta_1=min(1)$, and I got $|x^2|-|a^2|\le|x^2-a^2|<1$, adding |a^2| to both side of the inequality I got $|x^2|<1+|a^2|$, therefore $|x^2|+|a^2|<1+2|a^2|$ , and $|x^2+a^2|\le|x^2|+|a^2|<2|a^2|+1$

So I choose $\delta_1=min (1, \frac{\epsilon}{2|a^2|+1})$

Therefore I have, if $|x-a|<\delta$ , then $|x^2-a^2|<min (1, \frac{\epsilon}{2|a^2|+1})$

from my previous experience with $\lim_{x \to a} x^2$ , I know that I can pick $\delta=min(1, \frac{\epsilon}{2|a|+1})$, under these conditions

if $|x^2-a^2|<\epsilon$, then $|x-a|<\delta$

so if I let $min(1, \frac{\epsilon}{2|a^2|+1})$ play the role of $\epsilon$,

and I got $\delta=min(1, \frac{min(1,\frac{\epsilon}{2|a^2|+1})}{2|a|+1})$

when I check the solution, it seems like I got the right idea but the solution assume $\delta=min(1, \frac{min(1, \frac{\epsilon}{2(|a^2|+1)})}{2(|a|+1)})$

for some reason, I suppose I miss out a factor of two somewhere, but I check it again and again and could not find where I make a mistake, can somebody give me a clue?

2. Aug 18, 2016

### Staff: Mentor

Instead of factoring $x^4 - a^4 = (x^2 - a^2)(x^2 + a^4)$, look at the right side of this equation as $(x - a)(x + a)(x^2 + a^2) = (x - a)(x^3 + ax^2 + a^2x + a^3)$

3. Aug 18, 2016

### Aliax3012

Second Attempt at the Solution,

writing givens as the definition of limits

if $|x-a|<\delta$ then $|x^4-a^4|<\epsilon$

factoring out the right side of the condition

$|(x-1)(x^3+a^2x+ax^2+a^3)|<\epsilon$

choose $\delta=min(1)$ then I got, $|x-a|<1$, using triangle inequality I got $|x|-|a|\le|x-a|<1$, set $|x|-|a|<1$, and I got $|x|<1+|a|$

rewrite the expression $|(x^3+a^2x+ax^2+a^3)|\le|x^3|+|a^2|\times|x|+|a|\times|x^2|+|a^3|$

Now, I sub $|x|$ in, and the final result is $|(1+|a|)|^3+|a^2|\times|(1+|a|)|+|a|\times|(1+|a|)|^2+|a^3||$

now, I just choose that $\delta=min(1, \frac{\epsilon}{|(1+|a|)|^3+|a^2|\times|(1+|a|)|+|a|\times|(1+|a|)|^2+|a^3|})$

When I check the result, it matched up with the solution the book has given out in this second method, I'm just curious on why I got a different answer in first method that I choose to find $\delta$, it seems that I got the right methodology in the first one but just different answer and I don't know where I make my mistake, I don't know where the book got $2(|a|+1)$ and $2(|a^2|+1)$ instead of $2|a|+1$ and $2|a^2|+1$, I check again and again to see where I miss the 1 but I just can't find it, Thank you for your time Mark