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Math Help - Limit thread number 2?

  1. #1
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    Limit thread number 2?

    Can anyone help me with these two limit not using l hospital rule/.

    1. As x approaches 1

    (1/x)-1/(x-1)

    2. as h approaches zero

    (2h^3-h^2+5h)/(h)
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  2. #2
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    Re: Limit thread number 2?

    Quote Originally Posted by homeylova223 View Post

    2. as h approaches zero

    (2h^3-h^2+5h)/(h)
    Factor 'h' out of the numerator. What do you get?
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  3. #3
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    Re: Limit thread number 2?

    I get h(2h^2-h+5)/(h)
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    Re: Limit thread number 2?

    Cancel out h from numerator and denominator then make h=0
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    Re: Limit thread number 2?

    I got h=5 I was stuck but you make it seem so easy.
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    Re: Limit thread number 2?

    Quote Originally Posted by homeylova223 View Post
    I got h=5
    Sounds like a good answer

    Quote Originally Posted by homeylova223 View Post
    I was stuck but you make it seem so easy.
    The best way to master math is send lots of time studying. There's no such thing as too much math.
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  7. #7
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    Re: Limit thread number 2?

    On my first one (1/x)-1/(x-1) as x approaches 1

    Would It the final answer be zero because if I plug in one that is all I seem to get.
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    Re: Limit thread number 2?

    Nope, that is not the answer.

    Your function can also be written as -1/x(x-1)

    Have you tried to graph the function?
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    Re: Limit thread number 2?

    Quote Originally Posted by homeylova223 View Post
    I got h=5 I was stuck but you make it seem so easy.
    Not h= 5, the limit, as h goes to 0, is 5.

    For the first one, what you wrote, (1/x)-1/(x-1), does NOT have a limit as x goes to 1, but I think what you meant was ((1/x)- 1)/(x- 1).

    To find that limit, multiply both numerator and denominator by x to get (1- x)/(x(x-1)= -(x-1)/(x(x-1)). Now the x-1 terms cancel.
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