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Calculus: Limits help please

  1. Feb 22, 2005 #1
    1) [tex] \lim_{x\rightarrow 0}\frac {\sqrt[3]{1+x^2} -1}{x^2} [/tex]

    I tried doing a difference of cubes to the top and I got: [tex] \frac{\sqrt[3]{1+x^2}-1 ((\sqrt[3]{1+x^2})^2 + \sqrt[3]{1+x^2} + 1)} {x^2} [/tex]

    I know you need to get rid of the [tex] x^2 [/tex] on the bottom of the equation but now I'm stuck. Am I approaching this the right way and does anyone have any suggestions? Please and thank you.
    Last edited: Feb 22, 2005
  2. jcsd
  3. Feb 22, 2005 #2


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    L'Hospital's Rule applies:

    [tex] :(1): \ \ \ \ \lim_{x\rightarrow 0}\frac {\sqrt[3]{1+x^2} -1}{x^2} [/tex]

    [tex] :(2): \ \ \ \ = \ \lim_{x\rightarrow 0}\frac {(1/3)(2x)(1+x^2)^{-2/3}} {2x} [/tex]

    [tex] :(3): \ \ \ \ = \ \lim_{x\rightarrow 0} {(1/3)(1+x^2)^{-2/3}} [/tex]

    [tex] :(4): \ \ \ \ = \ (1/3) [/tex]

  4. Feb 23, 2005 #3
    you can use binomial expansion for

  5. Feb 23, 2005 #4


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    Make the subsitution [itex]h = x^2[/itex].

    [tex]\lim_{h\rightarrow 0}\frac {\sqrt[3]{1+h} -1}{h} [/tex]

    Does the form of the limit remind you of something familiar?
    Looks like the derivative of a function at some point...

    Alternatively, you could use the sub: [itex]t=\sqrt[3]{1+x^2}[/itex]
    Last edited: Feb 23, 2005
  6. Feb 23, 2005 #5

    I thought I got it but I didn't. [tex]\lim_{h\rightarrow 0}\frac {\sqrt[3]{1+h} -1}{h} [/tex] looks like is from the formula [tex]\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h} [/tex]. Or perhaps not? Could anyone give me another hint or suggestion involving the question using the method above,replacing [tex] x^2 [/tex] with [tex] h [/tex] please? Thank you very much.
  7. Feb 23, 2005 #6
    You're on the right track. It does look a lot like that. Now what's f(x) and what's a? Note that [itex]f(a + h) = \sqrt[3]{1 + h} [/itex] and [itex] f(a) = 1[/itex].

  8. Feb 24, 2005 #7
    So working backwards I got [tex] {\sqrt[3] {x} } [/tex] . Using the formula, replacing x with a [tex] 1+h [/tex] then: [tex] \frac {\sqrt[3] {1+h} - \sqrt[3] 1} {h} [/tex] which is [tex] \lim_{h\rightarrow 0}\frac {\sqrt[3]{1+h} -1}{h} [/tex]. Going back to [tex]\lim_{h\rightarrow 0}{\sqrt[3] {x} } [/tex] and replacing x with 0 gives you 0. I don't think I'm seeing it yet. Do you replace the [tex] {\sqrt[3] {x} } [/tex] back some point into the equation?
  9. Feb 24, 2005 #8
    Nope. You know that the limit is the derivative of the function evaluated at the point a. f(x) indeed equals x1/3. I think you'll agree with me in saying that calculating the derivative of x1/3 is pretty easy, yes? All you have to do is evaluate the derivative of it at a.

    Have you figured out what a is?

  10. Feb 24, 2005 #9
    could a be 1?
  11. Feb 24, 2005 #10


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    Yes, a is 1.

    I like Galileo's method, it shows insight.
  12. Feb 24, 2005 #11
    so using Galieo's method, do i replace x with 1 so [tex]\sqrt[3]1=1 [/tex] ?
  13. Feb 24, 2005 #12


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    The method with the derivatives is interesting/ingenious,but the one using
    [tex] A^{3}-B^{3}=(A-B)(A^{2}+AB+B^{3}) [/tex]

    is simply elegant...

    So the OP needs to redo calculations...

  14. Feb 24, 2005 #13


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    The substitution [itex]t=\sqrt[3]{1+x^2}[/itex] is easiest I think.
    The limit becomes:

    [tex]\lim_{t\to 1}\frac{t-1}{t^3-1}[/tex]

    De derivative method is cute and works sometimes, but usually a limit of this type shows up exactly when you want to calculate the derivative, so you cannot assume you know what the derivative of [itex]\sqrt[3]{x}[/itex] is.
    The same is true for l'hospitals method.
  15. Feb 24, 2005 #14
    You need to differentiate f(x) before you set x to a. Remember that the equation is f'(x)|x=a equals the limit, not f(a).

    Last edited: Feb 24, 2005
  16. Feb 24, 2005 #15


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    It's more like:
    [tex] f'(a)=f'(x)|_{x=a} [/tex] and NOT [tex] f'(a)=[f(a)]' (=0) [/tex]

  17. Feb 24, 2005 #16
    Yes, of course. Must spend more time composing posts.

  18. Feb 24, 2005 #17

    So by using this method, taking a difference of cubes on the bottom, the limit as t approaches 1 is [tex] \frac {1}{3} [/tex] which is the same answer as by using L'Hospital's Rule such as that in post #2. I'm sorry in advance for my stupidity but the bottom (which was originally [tex] x^2 [/tex] ) how did it become [tex] t^3-1 [/tex] ?
  19. Feb 24, 2005 #18


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    i dunno... but can't u just rationalize this limit?

    - Tu
  20. Feb 24, 2005 #19
    Solve for x2 in the equation [itex] t = \sqrt[3]{1+x^2} [/itex] and substitute it into the denominator.

  21. Feb 25, 2005 #20


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    Since the answer is known anyway:

    if [itex]t=\sqrt[3]{x^2+1}[/itex], then [itex]x^2=t^3-1[/itex].

    For the derivative method.
    After the substitution [itex]h=x^2[/itex] the limit becomes:
    [tex]\lim_{h \to 0}\frac{\sqrt[3]{1+h}-1}{h}[/tex]

    Now the derivative of [itex]\sqrt[3]{x}[/itex] at the point x=1 is:
    [tex]\lim_{h \to 0}\frac{\sqrt[3]{1+h}-\sqrt[3]{1}}{h}[/tex]
    which is exactly you limit.
    So the limit is equal to the value of the derivative of [itex]\sqrt[3]{x}[/itex] at the point x=1. The derivative is: [itex]\frac{1}{3\sqrt[3]{x}^{2}}[/itex], so at x=1 it's 1/3.
    (Seeing it as the derivative of [itex]\sqrt[3]{x+1}[/itex] in the point x=0 works too).
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