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Evaluate the limit

  1. Nov 5, 2004 #1

    quasar987

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    Can someone see how to do this? Here's what I have so far

    Evaluate

    [tex]\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{x-1} [/tex]

    This is an indeterminate form [itex]\frac{0}{0}[/itex]. Let's multiply by the conjugate 2 times.

    [tex]\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{x-1} = \lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{x-1} \frac{\sqrt[3]{x}+1}{\sqrt[3]{x}+1}\frac{\sqrt[3]{x}+1}{\sqrt[3]{x}+1} = \lim_{x\rightarrow 1}\frac{(x-1)+\sqrt[3]{x^2}-\sqrt[3]{x}}{(x-1)(\sqrt[3]{x}+1)^2}[/tex]

    I have tried going farther, setting [itex]y = \sqrt[3]{x}[/itex] but it's not coming to anything.
     
  2. jcsd
  3. Nov 5, 2004 #2

    Galileo

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    Making the substitution [itex]y = \sqrt[3]{x}[/itex] is the right idea.

    Notice y=1 is a root of [itex]y^3-1[/itex]. See if you can factor out (y-1).
    It'll give an expression that is probably familiar.
     
  4. Nov 5, 2004 #3

    quasar987

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    In terms of y, it's

    [tex]\frac{y^3+y^2-y-1}{(y^3-1)(y+1)^2}[/tex]

    or

    [tex]\frac{(y^2-1)(y+1)}{(y^3-1)(y+1)^2} = \frac{(y^2-1)}{(y^3-1)(y+1)}[/tex]

    I really don't see how to progress from there. :confused:
     
  5. Nov 5, 2004 #4
    Don't bother rationalizing it. Just make the substitution straight off.

    [tex]\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{x-1} = \lim_{y \to 1} \frac{y - 1}{y^3 - 1}[/tex]

    --Justin
     
  6. Nov 5, 2004 #5

    quasar987

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    I'm afraid this isn't inspiring me much more.
     
  7. Nov 5, 2004 #6
    Can you factor [itex]y^3 - 1[/itex] such that something cancels? Heck, there's only one way to factor [itex]y^3 - 1[/itex] either way, so why don't you give it a shot?

    After you factor and cancel, does the indeterminance issue remain?

    --Justin
     
  8. Nov 5, 2004 #7

    quasar987

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    Sorry, I was sure I had tried that before but that it led to nothing. Actually I had but I'd made a mistake in the division. Anyway, thanks a lot.
     
  9. Nov 5, 2004 #8
    From your post #3,
    [tex]\frac{(y^2-1)(y+1)}{(y^3-1)(y+1)^2} = \frac{(y^2-1)}{(y^3-1)(y+1)}[/tex]

    [tex]=\frac{(y+1)(y-1)}{(y-1)(y^2+y+1)(y+1)}[/tex]
     
  10. Nov 5, 2004 #9

    quasar987

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    :smile:




    (I have to type something)
     
  11. Nov 5, 2004 #10
    Use L'Hopital's Rule

    No, no, no! Use L'Hopital's Rule. Whenever a limit is such that an expression becomes an indeterminate form, you should use L'Hopital's Rule, which states that:

    [tex]f(a)=g(a)=0\Rightarrow\lim_{x\rightarrow a}\frac{f(x)}{g(x)} = \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}[/tex]

    What this means is that if you do get an indeterminate form, just differentiate the top and bottom of the fraction, and then try the value of the limit. If it is still an indeterminate form, just keep on differentiating (but this works only if it is an indeterminate form) and trying the value of the limit.

    In your case

    [tex]\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{x-1} = \lim_{x\rightarrow 1}3x^{-2/3} = 3(1)^{-2/3} = \frac{1}{3}[/tex]

    which is indeed the correct answer. I am too a university physics student ("college" in America), but I learnt this at what we call college (which is basically the course of study before you start university).

    Any further questions?

    Masud.
     
  12. Nov 5, 2004 #11

    quasar987

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    I thank you for pointing that out but I should have mentionned that I was not allowed to use L'Hospital's rule yet. This is for an analysis class, not a calculus class, and we haven't gotten to derivatives yet.

    I'm sure I'll soon have a lot of further questions though. :wink:

    I'll post them here in the next few days if you'd like to help.
     
    Last edited: Nov 5, 2004
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