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Homework Help: Need help with epsilon delta proof of f(x)=x^4+(1/x) as x goes to 1

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine the limit l for a given a and prove that it is the limit by showing how to find δ such that |f(x)-l|<ε for all x satisfying 0<|x-a|<δ.

    f(x)=[itex]x^{4}+\frac{1}{x}[/itex], a=1.

    2. Relevant equations
    I claim that [itex]\lim\limits_{x\rightarrow 1}x^{4}+\frac{1}{x}=2[/itex].

    3. The attempt at a solution
    I tried things like solving for x but all my attempts were fruitless. Could anyone point me in the right direction?
  2. jcsd
  3. Sep 10, 2012 #2
    |f(x) - L| = \left|x^4+\frac{1}{x} - 2\right| < \epsilon
    How can you make the above expression look like [itex]|x-1| < f(x)\epsilon[/itex]? Where f(x) is some expression.
  4. Sep 10, 2012 #3
    I thought I could factor it but I wasn't able to.

    Can I say, let [itex]\delta_{1}=1[/itex]. Then [itex]|x-1|<1\Rightarrow-1<x-1<1\Rightarrow0<x<1[/itex]. Then [itex]0<x^{4}<16[/itex]. We then also have [itex]1/2<\frac{1}{x}[/itex].

    That feels like it is a dead end not to mention that it doesn't look like itll become |x-1| anytime soon.
  5. Sep 10, 2012 #4


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    Homework Helper

    You need to use this definition :

    [itex]\forallε>0, \existsδ>0 | 0 < |x-a| < δ \Rightarrow |f(x) - L| < ε[/itex]

    You want to find a delta that will satisfy this definition. The trick with these sorts of problems is to massage |f(x)-L| until you are in the form |x-a| < δ.

    [itex]|f(x) - L| = \left|x^4+\frac{1}{x} - 2\right| = |\frac{x^5-2x+1}{x}| = |x-1||x^4+x^3+x^2+x-1|\frac{1}{|x|}[/itex]

    Now ask yourself... using the definition, what can you do to manipulate this expression into a suitable δ.

    EDIT : Also the triangle inequality will be useful here : |a+b| ≤ |a| + |b|
  6. Sep 10, 2012 #5
    Is there a certain way to factor out functions like that? I couldn't for the life of me figure out how to factor it...
  7. Sep 10, 2012 #6


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    I cheated a bit since i know that the |x-a| factor is ALWAYS going to show up when you massage your |f(x)-L| expression. I used polynomial long division coupled with the factor & remainder theorems.
  8. Sep 10, 2012 #7
    Looks like its time to brush up on long division.... Thanks a lot! I think I can take it from here.
  9. Sep 10, 2012 #8
    Ok. Here is my attempt at the proof.

    Proof: Suppose we are given ε>0. Then choose δ=min(1,[itex]\frac{2\epsilon}{31})[/itex].
    Thus we have
    0<|x-1|<δ &\Rightarrow|x-a|<\frac{2\epsilon}{31}.

    Now we will set [itex]|x-1|<1\Rightarrow-1\leq x-1\leq 1\Rightarrow 0<x<2.[/itex] Note that [itex]|x-1|\leq|x|+|-1|=|x|+1[/itex]. Then since |x|<2, we have [itex]|x|+1<2+1=3.[/itex] Thus [itex]|x-1|<3[/itex]/
    Note that [itex]|x^{4}+x^{3}+x^{2}+x-1|\leq|x^{4}+x^{3}+x^{2}|+|x-1|<|x^{4}+x^{3}+x^{2}|+3\leq|x^{4}+x^{3}|+|x^{2}|+3.[/itex] Now since |x|<2, we have [itex]|x^{2}|<2^{2}=4[/itex] so [itex]|x^{4}+x^{3}|+|x^{2}|+3<|x^{4}+x^{3}|+4+3=|x^{4}+x^{3}|+7\leq|x^{4}|+|x^{3}|+7[/itex]. Now [itex]|x^{4}|<2^{4}=16[/itex] and [itex]|x^{3}|<2^{3}=8[/itex] so [itex]|x^{4}|+|x^{3}|+7<16+8+7=31[/itex]. Thus [itex]|x^{4}+x^{3}+x^{2}+x-1|<31[/itex].

    Then it follows that
    |x-a|<\frac{2\epsilon}{31} &\Rightarrow\frac{1}{|x|}|x-1||x^{4}+x^{3}+x^{2}+x-1|<\frac{2\cdot 31\cdot\epsilon}{2\cdot 31}\\
    &\Rightarrow \frac{1}{|x|}|x-1||x^{4}+x^{3}+x^{2}+x-1|<\epsilon\\
    &\Rightarrow \frac{|x-1||x^{4}+x^{3}+x^{2}+x-1|}{|x|}<\epsilon\\
    &\Rightarrow \frac{|x^{5}-2x+1|}{|x|}<\epsilon\\
    &\Rightarrow |x^{4}+\frac{1}{x}-2|<\epsilon.
    This completes the proof.

    How does this look?
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