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Math Help - a simple limit problem.....

  1. #1
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    a simple limit problem.....

    Hey everyone.

    Somehow I do not know how to do this? I keep getting the wrong answers. If someone can show me or tell me what steps to take that would be great!

    also please try to do it using limits instead of derivatives. On the exams, the teacher will take off points if we use derivative methods.








    and




    and

    The horizontal asymptotes of the curve




    Thanks!!



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  2. #2
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    Oh I got the horizontal asymptotes. it is -18 and 18. now for the other two problems...
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  3. #3
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    Quote Originally Posted by makeupgirl1107 View Post
    Hey everyone.

    Somehow I do not know how to do this? I keep getting the wrong answers. If someone can show me or tell me what steps to take that would be great!

    also please try to do it using limits instead of derivatives. On the exams, the teacher will take off points if we use derivative methods.




    The first thing one would try is to just put x= 9/8. Of course, that doesn't work here (it never does for interesting problems) but it shows where the problem is- the denominator becomes 0 while the numerator does not. Now, what happens if x is very close to 9/8 but just slightly less? You can answer that by actually putting in such a number. 9/8= 1.125 so what happens if x= 1.120? In that case, the numerator is -22x= -24.26 and the denominator is 9- 8(1.120)= 9- 8.96= .04. -24.26/.04= -606.5. In other words, the fraction itself is a very large negative number- since the denominator is going to 0 and the numerator isn't, the limit will be "negative infinity". You could also do that by recognizing that 9- 8x= -8\left(x- \frac{9}{8}\right). If x is less than 9/8, the quantity in the parentheses, x- \frac{9}{8} is negative so -8\left(x- \frac{9}{8}\right) is the product of two negative numbers and so is negative. Since the numerator, -22x, is also negative, the fraction is negative and the limit is negative infinity. For x greater than 9/8, x- \frac{9}{8} is positive so that the denominator is negative and the fraction is the quotient of two negative numbers. The fraction is positive so the limit is positive infinity.

    and



    For \sqrt{x^2+ 4x+ 1}- x multiply by the fraction \frac{\sqrt{x^2+ 4x+ 1}+ x}{\sqrt{x^2+ 4x+ 1}+ x} then do thesame thing.


    and

    The horizontal asymptotes of the curve

    The horizontal asymptotes of the curve are, of course, the values of y as x goes to plus or minus infinity. I recommend that you divide both numerator and denominator by x. That way the numerator is the constant 18. What happens in the denominator?

    Thanks!!

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