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Evaluating a limit as x tends to ## \infty##

  1. May 13, 2015 #1
    1. The problem statement, all variables and given/known data

    $$ \lim_{x\to\infty} (\sqrt{a^2x^2 +bx + x} -ax) = $$
    b/2a
    b/a
    0
    2b/a
    2. Relevant equations
    Lim x tends to infinity 1/x = 0

    3. The attempt at a solution
    Taking a2x2 common from square root, we get
    $$ \lim_{x\to\infty} (ax \sqrt{1+\frac{b}{a^2x}+ \frac{1}{a^2x}} -ax) = $$
    Putting x as infinity we get,
    ax - ax = 0
    But Wolfram is showing something else
     
  2. jcsd
  3. May 13, 2015 #2

    ehild

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    Wolfram is right. You multiply something that tends to zero with something that goes to infinity. You can not conclude that the limit is zero.
     
  4. May 13, 2015 #3

    ehild

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    Use the approximation ##\sqrt{1+\delta} = 1+0.5 \delta ## valid if δ<<1.
     
  5. May 13, 2015 #4

    Mark44

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    You took the limit too soon. Instead, notice that both terms have a factor of ax, so factor that expression out so that you're taking the limit of ax * <other stuff>. What do you get now for your limit?
     
  6. May 13, 2015 #5

    SammyS

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    What is the result if a is not a positive number ?
     
  7. May 13, 2015 #6

    Ray Vickson

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    None of the possible answers listed is correct. What do you get if a > 0? What do you get if a < 0?
     
  8. May 13, 2015 #7
    I get

    $$ \lim_{x\to\infty} (ax( \sqrt{1+\frac{b}{a^2x}+ \frac{1}{a^2x}} -1)) $$
    Now what should I do to solve it?
    Can some formula, L Hopital rule etc. be applied here?
    Don't know for both the cases.
    It is amazing that most of the questions which I post are wrong or options are wrong. ( Test makers should verify the questions as it wastes time in exam hour.:mad:)
    Okay leaving the options,
    Can we say for this problem that the limit not exists?
     
  9. May 13, 2015 #8

    Mark44

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    ##\sqrt{a^2x^2} = |ax|##
    Since the limit is as ##x \to \infty##, we can assume that x > 0, but since no information is given about a, the best we can do is to write ##\sqrt{a^2x^2} = |a|x##.

    Regarding the limit above, use the fact that ##\lim AB = \lim A \cdot \lim B##, provided that both limits exist.
     
  10. May 13, 2015 #9
    So lim B = 0
    Lim A = ∞
    LimA.LimB = ∞ * 0 which is indeterminate.
     
  11. May 13, 2015 #10

    SammyS

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    I suspect the problem should have been ##\displaystyle\ \lim_{x\to\infty} (\sqrt{a^2x^2 +bx + c\ } -ax) \ ## or similar.

    Multiply & divide by the complex conjugate ( ← Added in quick Edit. Mark beat me to it!)
     
  12. May 13, 2015 #11

    Mark44

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    You're right. A trick that often works in cases like this is to multiply by the conjugate over itself. (IOW, multiply by 1 in a certain form).

    Going back to the original problem, ## \lim_{x\to\infty} (\sqrt{a^2x^2 +bx + x} -ax)##, try multiplying by ##\frac{\sqrt{a^2x^2 +bx + x} + ax}{\sqrt{a^2x^2 +bx + x} + ax}##

    BTW, are you sure about the quantity in the radical? It seems odd to me to have bx + x in there. I'm wondering if the last two terms should be bx + c. Just checking.

    Edit: Sammy beat me to it.
     
  13. May 13, 2015 #12

    Ray Vickson

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    No, you cannot say that. We have
    [tex] \sqrt{a^2 x^2 + bx + x} = |a x| \left( 1 + \frac{b+1}{a^2} \frac{1}{x} \right)^{1/2} - ax \doteq |a|x \left(1 + \frac{b+1}{2a^2} \frac{1}{x} \right) -ax\\
    = |a|x - ax + |a| \frac{b+1}{2a^2}.[/tex]
    for large ##x > 0##. Again, I ask: what do you get if ##a > 0##? What do you get if ##a < 0##?

    Note added in edit: it looks like one of your choices is correct IF a > 0 AND you mean bx + c inside the square root, instead of the bx + x that you wrote.



     
    Last edited: May 13, 2015
  14. May 13, 2015 #13
    In the radical there is bx + x term as I have typed earlier. It is sure.
    When I do the multiplying and dividing by conjugate and do some solving I get the the answer (b+1)/2a
    for a>0. For a<0 it is undefined.

    For bx + c term I am getting answer b/2a for a>0
    Why wolfram is showing something else?
     
  15. May 13, 2015 #14

    SammyS

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    It is almost surely a typo -- maybe not your typo, but somebody's typo. (I looked at my keyboard, and right beside the "x" key is the "c" key.)

    Tell Wolfram that a>0, (place ", a>0 " immediately after the rest of your input to Wolfram.)
     
  16. May 13, 2015 #15
    Getting not the result, see
    wolfram1
    And
    wolfram2
     
  17. May 13, 2015 #16

    SammyS

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    The following worked for me: Wolfman .

    , a>0

    You need the comma .
     
  18. May 13, 2015 #17
    Got it, thanks to all of you- ehild, Sammy,Ray and @Mark44 ( I don't know if it was a coincidence or you are really a fan of Mark 44 ?)
     
    Last edited: May 13, 2015
  19. May 13, 2015 #18

    Mark44

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    Purely a coincidence - I had never heard of Mark 44 before.
     
  20. May 13, 2015 #19

    BvU

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    Neither have I
    Mark 44 seems a fairly modern contraption. Mark44 is probably in existence since 1944 !
     
  21. May 13, 2015 #20

    Mark44

    Staff: Mentor

    Correctamundo!
     
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