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Math Help - Limit at infinity equaling infinity?

  1. #1
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    Limit at infinity equaling infinity?

    Hi guys,

    I'm doing limit exercises and I got this question:

    Evaluate the following limit:

    lim
    x → −∞ √(x2 + 5 x + 1) - x

    After working it out by multiplying by conjugate I get 5/0 which is undefined, but I know the answer is infinity. So where did I go wrong? This is the last step which I deduced 5/0 from:

    (5 + (0))/-sqrt(1 + (0) + (0)) + 1

    Any clarification is appreciated
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  2. #2
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    Hello, DannyMath!

    \displaystyle \text{Evaluate: }\;\lim_{x\to-\infty}\sqrt{x^2+5x+1} - x

    \text{. . . but I know the answer is infinity.} .?

    Multiply numerator and denominator by the conjugate:

    . . \displaystyle \frac{\sqrt{x^2+5x+1} - x}{1}\cdot\frac{\sqrt{x^2+5x+1} + x}{\sqrt{x^2+5x+1} + x}

    . . \displaystyle =\;\frac{(x^2+5x +1) - x^2}{\sqrt{x^2+5x+1} + x}  \;=\; \frac{5x+1}{\sqrt{x^2+5x+1} + x}


    \displaystyle \text{Divide numerator and denominator by }x\!:\;\; \displaystyle \frac{\dfrac{5x+1}{x}}{\dfrac{\sqrt{x^2+5x+1} + x}{x}}\;\;\bf[1]


    \displaystyle \text{The numerator is: }\;\frac{5x}{x} + \frac{1}{x} \;=\;5 + \frac{1}{x}


    \displaystyle \text{The denominator is: }\;\frac{\sqrt{x^2+5x+1}}{x} + \frac{x}{x}

    . . \displaystyle \;=\;\frac{\sqrt{x^2+5x+1}}{\sqrt{x^2}} + 1\;=\;\sqrt{\frac{x^2+5x+1}{x^2}} + 1

    . . \displaystyle =\;\sqrt{\frac{x^2}{x^2} + \frac{5x}{x^2} + \frac{1}{x^2}} + 1 \;=\; \sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1


    \displaystyle \text{Then }{\bf[1]}\text{ becomes: }\; \frac{5 + \frac{1}{x}}{\sqrt{1+\frac{5}{x} + \frac{1}{x^2}} + 1}


    \displaystyle \text{Therefore: }\;\lim_{x\to-\infty}\frac{5+\frac{1}{x}}{\sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1} \;=\;\frac{5 + 0}{\sqrt{1+0+0} + 1} \;=\;\frac{5}{2}

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  3. #3
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    Hey, thanks for the response! This was actually a 2 part exercise I was given for homework and I only posted the part (b). Part (a) was the exact same question but x goes to infinity rather than negative infinity. For that one I got 5/2 and the system told me I was right. But for the one I posted I got 5/0 but my graphing program showed it going to no particular horizontal asymptote so I had a sneaking suspicion that it was infinity. So that's what I submitted and it said I was correct :S ??

    Also, in the denominator, since you're turning x into sqrt(x^2), aren't you losing a negative sign if you don't write it -sqrt(x^2)? Since x would be negative as x >> infinity?
    Last edited by DannyMath; October 1st 2010 at 10:47 PM.
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, DannyMath!


    Multiply numerator and denominator by the conjugate:

    . . \displaystyle \frac{\sqrt{x^2+5x+1} - x}{1}\cdot\frac{\sqrt{x^2+5x+1} + x}{\sqrt{x^2+5x+1} + x}

    . . \displaystyle =\;\frac{(x^2+5x +1) - x^2}{\sqrt{x^2+5x+1} + x}  \;=\; \frac{5x+1}{\sqrt{x^2+5x+1} + x}


    \displaystyle \text{Divide numerator and denominator by }x\!:\;\; \displaystyle \frac{\dfrac{5x+1}{x}}{\dfrac{\sqrt{x^2+5x+1} + x}{x}}\;\;\bf[1]


    \displaystyle \text{The numerator is: }\;\frac{5x}{x} + \frac{1}{x} \;=\;5 + \frac{1}{x}


    \displaystyle \text{The denominator is: }\;\frac{\sqrt{x^2+5x+1}}{x} + \frac{x}{x}

    . . \displaystyle \;=\;\frac{\sqrt{x^2+5x+1}}{\sqrt{x^2}} + 1\;=\;\sqrt{\frac{x^2+5x+1}{x^2}} + 1


    You can't do the above since x<0...it can't go into the square root. What has to be done is:

    \frac{\sqrt{x^2+5x+1}}{x} + \frac{x}{x}=-\frac{\sqrt{x^2+5x+1}}{-x} + \frac{x}{x}=-\sqrt{\frac{x^2}{x^2} + \frac{5x}{x^2} + \frac{1}{x^2}} + 1 \;=-\; \sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1 ...etc.

    At the end the limit indeed is \infty

    Tonio



    . . \displaystyle =\;\sqrt{\frac{x^2}{x^2} + \frac{5x}{x^2} + \frac{1}{x^2}} + 1 \;=\; \sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1


    \displaystyle \text{Then }{\bf[1]}\text{ becomes: }\; \frac{5 + \frac{1}{x}}{\sqrt{1+\frac{5}{x} + \frac{1}{x^2}} + 1}


    \displaystyle \text{Therefore: }\;\lim_{x\to-\infty}\frac{5+\frac{1}{x}}{\sqrt{1 + \frac{5}{x} + \frac{1}{x^2}} + 1} \;=\;\frac{5 + 0}{\sqrt{1+0+0} + 1} \;=\;\frac{5}{2}

    .
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  5. #5
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    Thank you Tonio, I did indeed get that but doesn't that end up being (5 + (0))/-sqrt(1 + (0) + (0)) + 1 which turns into 5/0? I've just started limits so I don't know if that is the correct way to finish the problem (obviously it's not since it gives wrong answer :P)
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  6. #6
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    Quote Originally Posted by DannyMath View Post
    Thank you Tonio, I did indeed get that but doesn't that end up being (5 + (0))/-sqrt(1 + (0) + (0)) + 1 which turns into 5/0? I've just started limits so I don't know if that is the correct way to finish the problem (obviously it's not since it gives wrong answer :P)

    Well, as you know you can't divide by zero...

    What you actually get is an expression of the form \frac{c}{g(x)} , with c a constant and g(x) a

    function which converges to zero. It's not hard to see, even by means of the \epsilon,\,\delta definition, that the limit of

    this expression has as limit \pm\infty , depending on the constant's and the function's signs.

    Tonio
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  7. #7
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    Oh ok I will work on understanding that. We skipped the precise delta/epsilon description in our books and were just told to divide by greatest x degree in denominator. I just tried to go as far as I could with the definition. Maybe one day it will be as easy to see for me as it is for you guys! HAHA! Thanks!
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