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FP3 Limits Q

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    Q:

     \displaystyle\lim_{x\to \infty} \left( (x^2 + 3x)^\frac{1}{2} - x \right)

    My working thus far:

     \displaystyle \left( (x^2 + 3x)^\frac{1}{2} - x \right) = (3x)^\frac{1}{2} \times (1+ \frac {x}{3})^\frac{1}{2} - x

    then

     \displaystyle \sqrt {3x} \left ( 1 + \frac {x}{6} - \frac {x^2}{72} + \cdots \right) - x

    then

     \displaystyle  \left ( \sqrt {3x} + \frac {\sqrt {3} (x)^\frac {3}{2}}{6} - \frac {\sqrt {3} (x)^\frac {5}{2}}{72} + \cdots \right) - x

    and I am pretty much stuck from here.

    Thnx
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    (Original post by member910132)
    Q:

     \displaystyle lim_{x\to \infty} \left( (x^2 + 3x)^\frac{1}{2} - x \right)

    My working thus far:

     \displaystyle \left( (x^2 + 3x)^\frac{1}{2} - x \right) = (3x)^\frac{1}{2} \times (1+ \frac {x}{3})^\frac{1}{2} - x

    then

     \displaystyle \sqrt {3x} \left ( 1 + \frac {x}{6} - \frac {x^2}{72} + \cdots \right) - x

    then

     \displaystyle  \left ( \sqrt {3x} + \frac {\sqrt {3} (x)^\frac {3}{2}}{6} - \frac {\sqrt {3} (x)^\frac {5}{2}}{72} + \cdots \right) - x

    and I am pretty much stuck from here.

    Thnx
    I don't think that particular expansion helps a lot - in fact, it's not even valid as you're considering large x whilst it's only valid for |x|<3. I would start again and multiply the limit by \dfrac{\sqrt{x^2+3x} +x}{\sqrt{x^2+3x}+x}.
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    Trick: let x= 1/y.
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    (Original post by Farhan.Hanif93)
    I don't think that particular expansion helps a lot - in fact, it's not even valid as you're considering large x whilst it's only valid for |x|<3. I would start again and multiply the limit by \dfrac{\sqrt{x^2+3x} +x}{\sqrt{x^2+3x}+x}.
    But then I end up with



     \displaystyle \frac {3x}{\sqrt {x^2 +3x} + x}

    and I can't expand the bottom again because
     \displaystyle x \to \infty and the expansion won't be valid for that.
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    (Original post by f1mad)
    Trick: let x= 1/y.
    If I go ahead with this then does the limit change to
     \displaystyle \lim_{y \to 0} ?

    As then I can validly expand  \displaystyle( 1+ \frac {1}{3y})^2 ?
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    (Original post by member910132)
    If I go ahead with this then does the limit change to
     \displaystyle \lim_{y \to 0} ?
    Yes that becomes the limit.

    Then you get:

    (1/y^2 + 3/y)^1/2 -> (1/y^2(1+ 3y))^1/2

    ->((1/y)^2(1+ 3y))^1/2 -1/y it'll drop out fairly easily now.
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    (Original post by member910132)
    But then I end up with



     \displaystyle \frac {3x}{\sqrt {x^2 +3x} + x}

    and I can't expand the bottom again because
     \displaystyle x \to \infty and the expansion won't be valid for that.
    Note that x\equiv \dfrac{1}{1/x} and bring that new factor of \dfrac{1}{x} into the square root in the denominator. Things do simplify to a position to take the limit.

    If you wanted to use binomial expansion from the start, you could have noted \left(x^2+3x\right)^{\frac{1}{2}  } \equiv x\left(1 + \dfrac{3}{x}\right)^{\frac{1}{2}  }, which is valid for |x|>3 and hence will do the trick.
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    (Original post by Farhan.Hanif93)
    Note that x\equiv \dfrac{1}{1/x} and bring that new factor of \dfrac{1}{x} into the square root in the denominator. Things do simplify to a position to take the limit.

    If you wanted to use binomial expansion from the start, you could have noted \left(x^2+3x\right)^{\frac{1}{2}  } \equiv x\left(1 + \dfrac{3}{x}\right)^{\frac{1}{2}  }, which is valid for |x|>3 and hence will do the trick.
    If  \displaystyle -1 < \frac {3}{x} < 1 then

     \displaystyle -3 < x < 3 so how did you get

     \displaystyle |x| > 3 ?
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    (Original post by member910132)
    If  \displaystyle -1 < \frac {3}{x} < 1 then

     \displaystyle -3 < x < 3 so how did you get

     \displaystyle |x| > 3 ?
    When taking the reciprocal, you have to reverse the inequality signs.
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    (Original post by f1mad)
    Yes that becomes the limit.

    Then you get:

    (1/y^2 + 3/y)^1/2 -> (1/y^2(1+ 3y))^1/2

    ->((1/y)^2(1+ 3y))^1/2 -1/y it'll drop out fairly easily now.
    I just have a few concerns,

    This is my first time doing FP3 and so I don't have the mathematical intuition you guys have, so how on earth was I supposed to think of

     \displaystyle  x = \frac {1}{y}

    even after that, we get

     \displaystyle \left( \frac {1}{y^2} + \frac {3}{y})^\frac{1}{2}

    which is fine, but I could have turned that into either of

     \displaystyle (\frac {1}{3y})^\frac {1}{2} \times (1+ \frac{3}{y})^\frac{1}{2} = (\frac{1}{y^2})^\frac{1}{2} \times (1+3y)^\frac{1}{2}

    and you will only get the correct answer of  \displaystyle \frac{3}{2} if you chose the second one.

    So my point is how on earth was I supposed to think of all of that ? Is it likely the question would have at least said let  \displaystyle x = \frac{1}{y} ? F1, why dod you think of that substitution ? What made you think of it ?

    Secondly then, should I make a habit of remembering that if  \displaystyle x \to \infty then the expansions only valid for  \displaystyle -1< x<1 can't be used ? In fact, F1 is it this that made you use the substitution ? So you could justify the binomial series ?


    (Original post by Farhan.Hanif93)
    Note that x\equiv \dfrac{1}{1/x} and bring that new factor of \dfrac{1}{x} into the square root in the denominator. Things do simplify to a position to take the limit.

    If you wanted to use binomial expansion from the start, you could have noted \left(x^2+3x\right)^{\frac{1}{2}  } \equiv x\left(1 + \dfrac{3}{x}\right)^{\frac{1}{2}  }, which is valid for |x|>3 and hence will do the trick.
    Could you go into the "Note that x\equiv \dfrac{1}{1/x} and bring that new factor of \dfrac{1}{x} into the square root in the denominator." in more detail, I don't really understand ?
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    (Original post by f1mad)
    When taking the reciprocal, you have to reverse the inequality signs.
    So do the sings change when:

    1. taking the reciprocals

    2. multiplying/dividing by -ve number ?

    Any others I should be aware of ?

    Edit:  -1 < \frac {3}{x}<1

    -1>\frac{x}{3}>1 but how can something be smaller than -1 and bigger than 1 ?
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    (Original post by member910132)
    If  \displaystyle -1 < \frac {3}{x} < 1 then

     \displaystyle -3 < x < 3 so how did you get

     \displaystyle |x| > 3 ?
    \left|\dfrac{3}{x}\right|<1 \implies \left|\dfrac{x}{3}\right| >1

    (Original post by member910132)
    Could you go into the "Note that x\equiv \dfrac{1}{1/x} and bring that new factor of \dfrac{1}{x} into the square root in the denominator." in more detail, I don't really understand ?
    So we have \dfrac{3x}{\sqrt{x^2+3x} +x} \equiv \dfrac{3}{\frac{1}{x}(\sqrt{x^2+  3x} +x)} \equiv \dfrac{3}{\sqrt{1 + \frac{3}{x}} + 1}.

    Can you take the limit now?
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    (Original post by Farhan.Hanif93)
    \left|\dfrac{3}{x}\right|<1 \implies \left|\dfrac{x}{3}\right| >1

    Can you do that without the modulus sign ? Showing the both sides though.

    And how can  \frac{x}{3} be smaller than -1 and grater than 1?
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    (Original post by Farhan.Hanif93)

    If you wanted to use binomial expansion from the start, you could have noted \left(x^2+3x\right)^{\frac{1}{2}  } \equiv x\left(1 + \dfrac{3}{x}\right)^{\frac{1}{2}  }, which is valid for |x|>3 and hence will do the trick.
    Leaving aside this confusion of mine about the range of validity,

    after expanding it all I get to:

     \frac{3}{2x} - \frac{9}{8x^2} and that doesn't give me the answer of 3/2 as x -> infty.
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    (Original post by Farhan.Hanif93)
    \left|\dfrac{3}{x}\right|<1 \implies \left|\dfrac{x}{3}\right| >1


    So we have \dfrac{3x}{\sqrt{x^2+3x} +x} \equiv \dfrac{3}{\frac{1}{x}(\sqrt{x^2+  3x} +x)} \equiv \dfrac{3}{\sqrt{1 + \frac{3}{x}} + 1}.

    Can you take the limit now?
    Yea I got it form here, that is clever thinking though !!
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    (Original post by member910132)
    Leaving aside this confusion of mine about the range of validity,

    after expanding it all I get to:

     \frac{3}{2x} - \frac{9}{8x^2} and that doesn't give me the answer of 3/2 as x -> infty.
    That's not quite correct. It looks like you've forgotten about the factor of x on the outside of the binomial expansion. Once you multiply that in, the result follows.
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    Right, and to save me from making a brand new thread, can someone address the issue of flipping the inequality sign when taking reciprocals ?

     -1<x<1 \implies -1> \dfrac{1}{x} >1 but how can 1/x be bigger then 1 and smaller than -1 ?

    Or is it like this

     -1<x<1 \implies -1 > \frac{1}{x}\ \text{and} \ \frac{1}{x}>1 ?

    Edit: Would  |x| > 3 \ \text{be written as}\ -1 > \frac{1}{x}\ \text{and} \ \frac{1}{x}>1 and cannot it not be written as one inequality ?

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