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Math Help - [SOLVED] Some help with limits

  1. #1
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    [SOLVED] Some help with limits

    I have to evaluate the following limits:

    a) lim x -> 7
    (square root(x+2)) -3 / (x-7

    I was unable to solve this

    b) lim x -> 0
    1/x - 1/(x^2 + x)

    Somehow simplified it to 0 / x+1... this makes no sense.

    c) lim x -> -infinity
    (square root(4x^2 + 1)) / (3x-5)

    I somehow got denominator 0 and was unable to solve.

    What I did was take the x^2 out of the square root on top and got:

    x^2 square root(4+1/x^2) / 3/x - 5/x^2

    I am pretty sure the denominators go to zero and and this point I'm lost.


    Please help. I can do this and have done similar problems before, just need a little kick start.
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  2. #2
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    Quote Originally Posted by thekrown View Post
    I have to evaluate the following limits:

    a) lim x -> 7
    (square root(x+2)) -3 / (x-7

    I was unable to solve this

    rationalize the numerator ... multiply by \textcolor{red}{\frac{\sqrt{x+2} + 3}{\sqrt{x+2} + 3}}

    b) lim x -> 0
    1/x - 1/(x^2 + x)

    Somehow simplified it to 0 / x+1... this makes no sense.

    simplifies to \textcolor{red}{\frac{1}{x+1}}

    c) lim x -> -infinity
    (square root(4x^2 + 1)) / (3x-5)

    first, note that the limit will be less than 0. divide the numerator and denominator by x ...

    \textcolor{red}{\frac{\sqrt{4 + \frac{1}{x^2}}}{3 - \frac{5}{x}}}
    ...
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  3. #3
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    a) Okay gotcha, I can do this.

    b) I will redo this, I must of made a repetitive mistake. I will check my work again.

    c) I thought in this type of limit we would have to divided by the greatest power of the numerator, in this case x^2. How come just x?
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  4. #4
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    Quote Originally Posted by thekrown View Post
    a) Okay gotcha, I can do this.

    b) I will redo this, I must of made a repetitive mistake. I will check my work again.

    c) I thought in this type of limit we would have to divided by the greatest power of the numerator, in this case x^2. How come just x?
    if x goes under the radical, what does it become?
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  5. #5
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    Let me see if I understand this here... see my teacher uses a robotic-teaching method where he says "just do this" and it works. So i haven't learned the proper reasoning to this type of limit.

    If what I think you're saying is, the numerator will divide by the greatest power so our square root(4x^2+1) is divided by x^2 to get square root(4+1/x).

    Then for the denominator, we use the square root of x^2 and divide the numerator by that?

    Is that why it's just x?
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  6. #6
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    Quote Originally Posted by thekrown View Post
    Let me see if I understand this here... see my teacher uses a robotic-teaching method where he says "just do this" and it works. So i haven't learned the proper reasoning to this type of limit.

    If what I think you're saying is, the numerator will divide by the greatest power so our square root(4x^2+1) is divided by x^2 to get square root(4+1/x).

    Then for the denominator, we use the square root of x^2 and divide the numerator by that?

    Is that why it's just x?
    the full monty ...

    \frac{\sqrt{4x^2+1}}{|x|} = \frac{\sqrt{4x^2+1}}{\sqrt{x^2}} = \sqrt{\frac{4x^2+1}{x^2}}<br />

    now understand that you're really dividing by |x| since \sqrt{x^2} = |x|

    no effect on the numerator, since it is positive for all x.

    for the denominator, however ...

    \frac{3x-5}{|x|} = -3 - \frac{5}{|x|} since x < 0
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