- #1
Aliax3012
- 2
- 0
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. Homework Statement
find a ##\delta## such that ##|f(x)-L|<\epsilon## for all x satisfying ##0<|x-a|<\delta##
[/B]
##f(x)=x^4##; ##l=a^4##
[/B]
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?
1. Homework Statement
find a ##\delta## such that ##|f(x)-L|<\epsilon## for all x satisfying ##0<|x-a|<\delta##
Homework Equations
[/B]
##f(x)=x^4##; ##l=a^4##
The Attempt at a Solution
[/B]
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?