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Limit problem

  1. Feb 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Find [tex]lim_{x->- \infty} \; \frac{(x^6+8)^{1/3}}{4x^2+(3x^4+1)^{1/2}}[/tex]

    2. Relevant equations
    N/A

    3. The attempt at a solution
    Factoring out [tex]\frac {(-x^6)^{1/3}}{-x^2}[/tex] leaves me with [tex]\frac{(-1-8x^{-6})}{-4+(3+x^{-4})^{1/2
    }}[/tex] Taking the limit at infinity gives me [tex]\frac{-1}{-4+3^{1/2}}[/tex], which is wrong.
     
  2. jcsd
  3. Feb 5, 2015 #2

    Quantum Defect

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    Why not factor out (x^2/x^2) on the top?... i.e. lim x-> -inf [1+8/x^6]^(1/3) / [4 + SQRT(3+ 1/x^4)]
     
  4. Feb 5, 2015 #3

    Svein

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    Your arithmetic is wrong (in the denominator).
     
  5. Feb 5, 2015 #4

    Ray Vickson

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    Why not use ##(x^6+8)^{1/3} = (x^6)^{1/3} (1 + 8 x^{-6})^{1/3} \doteq x^2 [1 + (8/3)x^{-6} + O(x^{-12}) ]##, and a similar result for ##(3 x^4 + 1)^{1/2}##?
     
  6. Feb 5, 2015 #5
    Ah, you're right. it should be MINUS root 3 because [tex]\sqrt{x^4}=-x^2[/tex]

    I don't understand how you arrived at this part.

    If I factored out positive x^2, how would I evaluate x at - infinity? Wouldn't I just evaluate at positive infinity, which happens to produce the same result in this case?
     
  7. Feb 5, 2015 #6

    Quantum Defect

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    Factoring out +x^2 says nothing about what the sign of x is, does it?

    The function whose limit is being evaluated is an even function of x, no? f(-x) = f(x) Wouldn't they be expected to have the same limit for very large positive and very large negative values of x?
     
  8. Feb 5, 2015 #7

    Ray Vickson

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  9. Feb 5, 2015 #8
    Can you link me the formula for that expansion? I've never seen it before.
     
  10. Feb 5, 2015 #9

    Quantum Defect

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  11. Feb 5, 2015 #10
    Where did the O(v^2) come from? The formula seems to stop at 1+rv.
     
  12. Feb 5, 2015 #11

    Ray Vickson

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    Google "binomial expansion". Alternatively, just apply the Maclauren (Taylor) series around ##v=0## to the function ##f(v) = (1+v)^r##.
     
  13. Feb 5, 2015 #12
    I won't be learning about power series until the end of next semester, so it's likely that your explanation is a bit above my level for the time being.

    I'm still a bit confused by why it's - root 3.

    The only example problem in my book defines (x^6)^1/2= - x^3, but when they factor [tex]\sqrt{25x^6}[/tex] it comes out to +5 rather than -5.
     
  14. Feb 5, 2015 #13

    HallsofIvy

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    No, it doesn't. It leaves you with [tex]\frac{(-1-8x^{-6})}{-4- (3+x^{-4})^{1/2
    }}[/tex]

     
  15. Feb 5, 2015 #14
  16. Feb 5, 2015 #15

    Ray Vickson

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    We should NOT have
    [tex] \sqrt{x^4} = -x^2 \:\longleftarrow\;\text{wrong!}[/tex]
    For any real ##x## we have ##\sqrt{x^4} = x^2## (because ##\sqrt{x^4} > 0 ## and ## x^2 > 0## if ##x \neq 0##; this is the same, whether ##x## is positive or negative. However, ##\sqrt{x^6}## is different. Again, we always have ##\sqrt{x^6} > 0##, but now ##x^3## is >0 if ##x > 0## and is < 0 if ##x < 0##. In fact, if ##x < 0## then ##x^3 < 0##, so we need ##\sqrt{x^6} = -x^3## in that case (where ##x < 0##). On the other hand, if ##x > 0## then we have ##\sqrt{x^6} = x^3##. To summarize:
    [tex] \sqrt{x^4} = x^2 \;\; \text{always}\\
    \sqrt{x^6} = \begin{cases} x^3 & \text{if} \;\; x > 0 \\
    -x^3 & \text{if} \;\; x < 0
    \end{cases}
    [/tex]
     
    Last edited: Feb 6, 2015
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