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Math Help - Limits help

  1. #1
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    Exclamation Limits help

    1. Find the lim x goes to 4 on the postive side of the function x-5/(4-x^3)(x-7)^2.
    2. Find the lim x goes to postive Pi/2 of the function tan x.
    3. Find the lim x goes to infinity (2x+1)/x^2-(2x-1)/x^2.
    4. Find the lim x goes to infinity (x^2)/(2x+1) - (x^2)/(2x-1)
    5. Fiind the lim x goes to infinity x^3/3^x
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  2. #2
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    Quote Originally Posted by Nimmy
    1. Find the lim x goes to 4 on the postive side of the function x-5/(4-x^3)(x-7)^2.[/tex]
    It is countinous, simplfy substitute 4 for x
    2. Find the lim x goes to postive Pi/2 of the function tan x.
    The limits does not exist, however +\infty

    5. Fiind the lim x goes to infinity x^3/3^x
    Exponential always overtake polynomials so the limit is zero.
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  3. #3
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    Quote Originally Posted by Nimmy
    4. Find the lim x goes to infinity (x^2)/(2x+1) - (x^2)/(2x-1)

    <br />
\frac{x^2}{2x+1} - \frac{x^2}{2x-1}=\frac{x^2(2x-1)-x^2(2x+1))}{4x^2-1}=-\frac{2x}{4x^2-1} =\frac{2}{4+1/x^2}<br />

    Therefore:

    <br />
\lim_{x \to \infty} \frac{x^2}{2x+1} - \frac{x^2}{2x-1}= \lim_{x \to \infty}\frac{2}{4+1/x^2}=1/2<br />

    RonL
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  4. #4
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    Exclamation Math Help Homework

    Find the lim x goes to inifinity (2/3x-4)? and also for the equation of square root of 4x+1.
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  5. #5
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    Is the first one \lim_{x \rightarrow \infty} \frac{2}{3}x-4 or \lim_{x \rightarrow \infty}\frac{2}{3x}-4?
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  6. #6
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    Exclamation

    Ok nvr mind about that problem but i do need help on square root of 4x+1 using the def. of dervitaitve. Dont use the calculus way!!!
    Last edited by Nimmy; September 6th 2006 at 08:14 PM.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Nimmy
    Ok nvr mind about that problem but i do need help on square root of 4x+1 using the def. of dervitaitve. Dont use the calculus way!!!
    How can we not use Calculus if we are using the definition of a derivative??

    -Dan

    For the record:
    \frac{d}{dx}\sqrt{4x+1} = \frac{2}{\sqrt{4x+1}} which is positive for all x in its domain, so we see that \sqrt{4x+1} is monotonically increasing on  [ -\frac{1}{4}, \infty ) and it has no horizontal asymptote. Thus \lim_{x \to \infty} \sqrt{4x+1} \to \infty .
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Nimmy
    Find the lim x goes to inifinity (2/3x-4)? and also for the equation of square root of 4x+1.
    What about \sqrt{4x+1} is it that you want help with?

    RonL
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  9. #9
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    Exclamation Math Help Homework

    Find a and b so that the slope of the tangent line to f(x)= ax^2+bx (1,3) is -2.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Nimmy View Post
    Find a and b so that the slope of the tangent line to f(x)= ax^2+bx (1,3) is -2.
    When you are asking a new question, please start a new thread. It's much less confusing that way, and much easier for someone else to find help on a similar problem.

    The slope of the tangent line to f(x) = ax^2 + bx is the first derivative f'(x) = 2ax + b. At the point (1,3) f'(1) = 2a + b = -2. There are many possible solutions to this equation, so I'll simply say that you can pick any "a" value and the required "b" value will then be b = -2a - 2.

    -Dan
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  11. #11
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    Exclamation

    So do u basically plug in -2a-2 into f'(x) or f(x)?
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  12. #12
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    Question Answered.
    Thread Closed!
    If you are going to start asking new questions start a new thread.
    -=USER WARNED=-
    Infraction recieved.
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