1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Epsilon and delta of x^3

  1. May 30, 2017 #1
    1. The problem statement, all variables and given/known data
    find a δ for a given ε for f(x)=x3 around c=5:
    $$\vert x-5\vert<\delta~\Rightarrow~\vert x^3-5^3 \vert < \epsilon$$

    2. Relevant equations
    Continuity:
    $$\vert x-c \vert < \delta~\Rightarrow~\vert f(x)-f(c) \vert < \epsilon$$
    $$\delta=\delta(c,\epsilon)$$

    3. The attempt at a solution
    $$\vert x^3-5^3 \vert =\vert x-5 \vert \cdot \vert x^2+5x+25 \vert$$
    $$=\vert x-5 \vert \cdot \vert (x-5)^2+15(x-5)+75 \vert < \epsilon$$
    $$\delta \cdot (\delta^2+15\delta+75)<\epsilon$$
     
  2. jcsd
  3. May 30, 2017 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hint: look at ##x^2+5x+25##. What can you do with that if ##x## is "close enough" to ##5##?
     
  4. May 30, 2017 #3

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    A) What is your question?
    B) If I give you a value for epsilon, can you give me a value for delta (a value for delta itself, that is) ?
     
  5. May 30, 2017 #4
    If x is close to 5 from the left side then:
    $$\vert x^3-5^3 \vert =\vert x-5 \vert \cdot \vert x^2+5x+25 \vert < 75\delta~\rightarrow~\delta<\frac{\epsilon}{75}$$
    But if x is on the right?
     
  6. May 30, 2017 #5

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I was thinking more that if ##x## is close to ##5## then ##x^2 + 5x + 25 \approx 75##.

    And, if ##|x - 5| < 1##, say, then how big can ##x^2 + 5x + 25## be?
     
  7. May 30, 2017 #6
    if ##~|x-5|<1##:
    $$\vert x^3-5^3 \vert \triangleq\epsilon=\vert x-5 \vert \cdot \vert x^2+5x+25 \vert < 91\delta~\rightarrow~\delta>\frac{\epsilon}{91}$$
    But it should be ##~\delta<\frac{\epsilon}{91}##
     
  8. May 30, 2017 #7

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think you are getting yourself confused in some way. You start with:

    ##|x^3 - 5^3| = |x-5||x^2 +5x + 25|##

    Now you let ##|x -5| < 1##, which implies ##4 < x < 6##, which implies that ##|x^2 +5x + 25| < 91##

    Putting that together you have:

    ##|x -5| < 1 \ \Rightarrow \ |x^3 - 5^3| < 91|x-5|##

    Now, if you let ##\epsilon > 0## can you finish it off?

    Hint: Let ##|x-5| < \dots ##
     
  9. May 31, 2017 #8
    $$\delta\triangleq |x-5|=\frac{|x^3-5^3|}{|x^2+5x+25|}=\frac{\epsilon}{|x^2+5x+25|}<\frac{\epsilon}{91}$$
    $$\Rightarrow\delta<\frac{\epsilon}{91}$$
     
  10. May 31, 2017 #9

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I give you ##\epsilon = 200\ ##. Then ## \delta =2 < {200\over 91}\ ## is not good enough ...
     
  11. May 31, 2017 #10

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's the background analysis, but it's not a proof, because you've got no explanation of what you are doing and the implication is the wrong way at the end.

    What you've shown is:

    ##|x - 5| = \delta \ \Rightarrow \ \delta < \frac{\epsilon}{91}##

    (I'm sorry to say I don't know what the little triangle means.)

    You need to turn the logic round, so that for a given ##\epsilon## you have a well-defined ##\delta##.

    Also, don't forget that you needed ##|x-5| < 1## as well.
     
  12. May 31, 2017 #11
    I wasn't told, in the original question, that ##~|x-5| < 1~##.
    And for |x-5|<1 it's correct ##~\delta < \frac{\epsilon}{91}##. but for all real numbers on x i don't know what to do since |x-5| doesn't appear in (x2+5x+25)
    ##\triangleq## means definition, a symbol given to
     
  13. May 31, 2017 #12

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You weren't told that ##|x -5| < \epsilon /91## either. So, where did that come from?
     
  14. May 31, 2017 #13

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

     
  15. May 31, 2017 #14

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You weren't told that ##|x -5| < \epsilon /91## either.
     
  16. May 31, 2017 #15
    ##|x -5| < \epsilon /91~## came from the condition ##~|x - 5| < 1## you proposed, and i guess ##~\delta<\frac{\epsilon}{91}~## satisfies this condition, but i am not sure.
    And even if i solve correctly for ##~|x - 5| < 1~## it's not a general answer for all x
     
  17. May 31, 2017 #16

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes it is a solution. You are free to choose ##x## as close to 5 as you like. That's the whole point. If you can't choose ##x## close to ##5## how can you prove continuity?

    If you look at ##x = 1,000,000## then even large ##\epsilon## is a problem.
     
  18. May 31, 2017 #17
    I don't think it's a solution, since the definition of continuity first chooses ε and derives a δ. i choose ε as small as i want and the δ becomes small, x comes close to 5.
    Yes, large ε is out of the domain |x-5|<1 (ε is not in the domain, it's on the Y axis, of course)
     
  19. May 31, 2017 #18

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Unfortunately, then, I don't think I can help you any further.
     
  20. Jun 1, 2017 #19
    I am sorry i sounded not polite, PeroK, please help, i have no idea how to solve even for |x-5|<1, as you suggested and i am very frustrated
     
  21. Jun 1, 2017 #20

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You have shown that for ##\varepsilon < 91 \ ## the condition ##\delta < {\varepsilon \over 91} \ ## is sufficient. I think that should be good enough for this exercise.

    Personally, all I wanted to point out was that: just ##\delta < {\varepsilon \over 91} \ ## in itself is not really sufficient if you take the exercise wording literally.

    In analysis statements are made like 'For all ##\varepsilon > 0 \ ## there exists a ##\delta > 0\ ## such that ... ' with the silent implication/intention that ##\varepsilon## can be made as small as desired (you ##\ ## look at/worry about ##\ ## the possibility to find a ##\delta## for ##\varepsilon \downarrow 0## ).

    Outside of such a context nitpckers like me can point out that a simple expression for ##\delta## may fail for big ##\varepsilon##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted