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Math Help - limit as x approaches -infinity?

  1. #1
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    limit as x approaches -infinity?

    What is the limit as x approaches -inf of sqrt(x^2+5x+1) -x

    I've tried to do this using the conjugate and I got -5 but my answer sheer tells me thats not it...but I don't know how to get the right answer
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  2. #2
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    Quote Originally Posted by katekate View Post
    What is the limit as x approaches -inf of sqrt(x^2+5x+1) -x

    I've tried to do this using the conjugate and I got -5 but my answer sheer tells me thats not it...but I don't know how to get the right answer
    Is the expression right, we saw this a couple of days ago with a +x at the end.

    (the limit you give should obviously go to +inf)

    CB
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  3. #3
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    Quote Originally Posted by katekate View Post
    What is the limit as x approaches -inf of sqrt(x^2+5x+1) -x

    I've tried to do this using the conjugate and I got -5 but my answer sheer tells me thats not it...but I don't know how to get the right answer

    Of course the answer cannot negative: square root is alwasy positive, and the -x part is ALSO positive sicne x --> -oo ! ==> the limit, in case it exists and is finite, must be non-negative.

    Now, the conjugate thing is a good idea, and then we get:

    [x^2 + 5x + 1 - x^2]/[Sqrt(x^2+5x+1) + x] = [5x + 1]/[Sqrt(x^2+5x+1) + x]

    The trick now is to mutiply by (1/-x)/(1/-x)....why -1/x? Because we'll want to "introduce" 1/-x into the square, but as x --> -oo we get that for almost all the values of x it is negative and we need something POSITIVE inside a square root, so:

    {[5x + 1]/[Sqrt(x^2+5x+1) + x]}(1/-x)/(1/-x) = [-5 - 1/x]/[Sqrt(1 + 5/x + 1/x^2) - 1] --- pay attention to the fact that we're introducing 1/-x into the Square root BUT already inside it becomes 1/x^2, of course.


    We can now see that the numerator --> -5 whereas the denominator --> 0 (from the left since 1 + 5/x + 1/x^2 < 1...remember: x < 0) ==> the limit doesn't exist in finite form, and in fact the expression --> oo as x --> -oo

    Tonio
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  4. #4
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    That's definitely the right expression, ending with -x and to -inf
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  5. #5
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    try substituting really large negative numbers for x...the result are really large pos numbers
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  6. #6
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    Quote Originally Posted by katekate View Post
    That's definitely the right expression, ending with -x and to -inf
    So what does your answer sheet tell you the answer should be.

    CB
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