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  1. #1
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    slant asymptote

    show that the curve sqrt(x^2+4x) has two asymptotes: y=x+2 and y+-x-2.
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    Quote Originally Posted by johntuan View Post
    show that the curve sqrt(x^2+4x) has two asymptotes: y=x+2 and y+-x-2.
    A 'nasty' way (mathematicians turn away):

    y = \sqrt{(x + 2)^2 - 4} \rightarrow \sqrt{(x+2)^2} = \pm(x + 2).

    A much better way is to expand \sqrt{x^2+4x} using the generalised binomial theorem and then take the limit.
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    haha..but i dont understand how +-(x+2) becomes asymptotes?...the way i would get S.A is long division or the limit way...but this one is weird because its not a rational function
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    Quote Originally Posted by johntuan View Post
    haha..but i dont understand how +-(x+2) becomes asymptotes?...the way i would get S.A is long division or the limit way...but this one is weird because its not a rational function
    Slant asymptotes are lines that the curve approaches as x --> +oo or x --> -oo. As x --> +oo or -oo, (x+2)^2 - 4 \rightarrow (x+2)^2 and the square root of this is \pm (x+2). This is not a rigorous way of doing it. The mathematicians of MHF will pounce on this and make mince meat of it.

    But I'm not sure you're at a level where a rigorous approach would make much sense, so you'll just have to make do with the one above.
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    Re: slant asymptote

    Hi all. I was confused by this limit too. I am not quite sure what is wrong with this reasoning. I have a suspicion but I would like to see other thoughts.

    For x > 0, sqrt(x^2 + 4x) is the same as (x^2 + 4x)/sqrt(x^2 + 4x) = (x + 4)/sqrt(1 + 4/x).
    At this point, why can we just let 4/x ->0 as x -> inf and we get the limit tending to x + 4?
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  6. #6
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    Re: slant asymptote

    the graph of $y = \sqrt{x^2+4x}$ is the upper branch of a hyperbola with a horizontal transverse axis ...

    $y^2 = x^2+4x$

    $y^2+4 = (x+2)^2$

    $(x+2)^2 - y^2 = 4$

    note the standard form for a hyperbola is $\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1$

    $\dfrac{(x+2)^2}{2^2} - \dfrac{y^2}{2^2} = 1$ , a hyperbola centered at $(h,k) = (-2,0)$

    Asymptotes have slope $m = \pm \dfrac{b}{a} = \pm \dfrac{2}{2} = \pm 1$

    since the asymptotes pass through the hyperbola center ...

    $y - 0 = -1(x + 2) \implies y = -x - 2$

    and

    $y-0 = 1(x+2) \implies y = x+2$
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    Re: slant asymptote

    Quote Originally Posted by cjfogel View Post
    I am not quite sure what is wrong with this reasoning.
    The first thing I notice, is that it is not rigorous to state (x+2)^2-4 \to x+2. Instead we should be saying something like
    (x+2)^2-4 = (x+2)^2{(x+2)^2-4 \over (x+2)^2} = (x+2)^2\left(1 - {4 \over (x+2)^2}\right) \to (x+2)^2
    Then we get that
    y = \sqrt{x^2+4x} \to \sqrt{(x+2)^2} = \left|x+2\right|

    And the modulus signs are critical, because when x < -2 as it is when x \to -\infty, this gives us y \to -(x+2) = -x-2.
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    Re: slant asymptote

    OK. I realise you meant that you have some different reasoning that you know to be wrong.
    Quote Originally Posted by cjfogel View Post
    For x > 0, \sqrt{x^2 + 4x}= {x^2 + 4x \over \sqrt{x^2 + 4x}} = {x + 4 \over \sqrt{1 + \frac4x}}.
    At this point, why can we just let \tfrac4x \to 0 as x \to \infty and we get the limit tending to x + 4?
    The problem comes in how \sqrt{1+\frac4x} \to 1. The (Maclaurin) series expansion of \sqrt{1+y} about y=0 is \sqrt{1+y} = 1 + \tfrac12 y + \ldots so
    \begin{aligned}{x + 4 \over \sqrt{1 + \frac4x}} &= {(x+4)\sqrt{1-\frac4x} \over \sqrt{1+\frac4x}\sqrt{1-\frac4x}} \\ & \approx {(x+4)(1-\frac2x) \over 1 - \frac4x} = {x + 4 - 2 - \frac8x \over 1 - \frac4x} = {x+2 - \frac8x \over 1 -\frac4x} \\ &\approx x+2 \end{aligned}
    for large x.

    Alternatively using the Maclaurin series above and {1 \over 1+y} = 1 - y + \ldots, we have
    \begin{aligned}{x + 4 \over \sqrt{1 + \frac4x}} &\approx {x+4 \over 1 + \frac2x} \\ &\approx (x+4) (1 - \frac2x) = x + 4 - 2 - \frac8x \\ &\approx x+2\end{aligned}
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