More help needed manipulating terms

Discussion in 'Differentiation and Integration' started by mmesford, Jan 2, 2022.

  1. mmesford

    mmesford

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    My basic algebra skills are still letting me down. Trying to find the limit of this term:

    [(2x-3)(sqrt{x}-1)]/[2x^2+x-3]

    It seems obvious that I need to get the ‘x’ out of the sqrt but I can’t figure out how. I tried multiplying top and bottom by (sqrt{x}+1) but that just over the problem to the denominator. I tried squaring top and bottom but I’m left with (x-2*sqrt{x}+1) in the numerator. I can’t envision anything that frees up the sqrt{x} term. Any ideas?
     
    mmesford, Jan 2, 2022
    #1
  2. mmesford

    MathLover1

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    [​IMG]..........factor 2x^2+x-3= (x - 1) (2 x + 3)

    =[​IMG]........rationalize , multiply both numerator and denominator by (sqrt(x)+1)

    =[​IMG]

    =[​IMG]

    =[​IMG].......simplify

    =[​IMG]
     
    MathLover1, Jan 2, 2022
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  3. mmesford

    nycmathguy

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    Nicely-done.
     
    nycmathguy, Jan 2, 2022
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  4. mmesford

    mmesford

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    Thank you, but I don’t understand how this helps. I was able to get this far but don’t I need to get rid of the sqrt{1} in order to calculate the limit?
    If I plug 1 into the final form I end up with:
    [(2-3)(sqrt{1}-1)]/2+1-3 which simplifies to:
    -1/(5*sqrt{1}+5)
    If I ignore the sqrt{1} it simplifies to -1/10 which is the answer in the book. But how do I get away with ignoring the sqrt{1}?
     
    mmesford, Jan 3, 2022
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  5. mmesford

    nycmathguy

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    I have never taken calculus but I understand what she did. What is x tending to? If, for example, x approaches 0, no need to remove the square root from the denominator. Just replace x with whatever x is approaching and simplify.
     
    nycmathguy, Jan 3, 2022
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  6. mmesford

    mmesford

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    Sorry, I forgot to mention, we’re looking for the limit as x approaches 1. But you have the concept down.
     
    mmesford, Jan 3, 2022
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  7. mmesford

    nycmathguy

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    Beautiful. Replace every x you see with 1 and simplify.
     
    nycmathguy, Jan 3, 2022
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  8. mmesford

    nycmathguy

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    As x--->1

    [2(1) - 3]/[(2•1 + 3)(sqrt{1} + 1)]

    Simplify that to find your limit.
     
    nycmathguy, Jan 3, 2022
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  9. mmesford

    MathLover1

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    observe the graph

    [​IMG]



    if x=1-> then [​IMG]
     
    Last edited: Jan 3, 2022
    MathLover1, Jan 3, 2022
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  10. mmesford

    mmesford

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    So it’s legal to just ignore the imaginary part? And the only reason we went to all that trouble was so there wasn’t a zero in the denominator?
     
    mmesford, Jan 3, 2022
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  11. mmesford

    MathLover1

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    yes, and the only reason you can face trouble was that x=1 would make the denominator of the original expression equal to zero:

    [(2x-3)(sqrt{x}-1)]/[2x^2+x-3]=>if x=1 than [2*1^2+1-3]=2+1-3=0
     
    MathLover1, Jan 3, 2022
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  12. mmesford

    mmesford

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    Great, thanks!
     
    mmesford, Jan 3, 2022
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  13. mmesford

    nycmathguy

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    Is my reply to this member correct?
     
    nycmathguy, Jan 4, 2022
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  14. mmesford

    MathLover1

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    yes
     
    MathLover1, Jan 4, 2022
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  15. mmesford

    nycmathguy

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    Cool. Just posted two more Half-Angle Formulas: 41 and 44.
     
    nycmathguy, Jan 4, 2022
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  16. mmesford

    Country Boy

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    Not that in lim_(x->a) f(x)/g(x),
    if neither f(a) nor g(a) is 0, the limit is f(a)/g(a).
    if f(a)= 0 but g(a) is not 0, the limit is 0.
    if f(a) is not 0 but g(a)= 0, the limit does not exist.

    The only 'problem' is if f(a)= 0 and g(a)= 0. If f and g are polynomials (or can be made into polynomials as here) then they must both have factors of (x- a) which can be canceled.
     
    Country Boy, Jan 5, 2022
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