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Math Help - help would be very much appreciated please

  1. #1
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    Thumbs down help would be very much appreciated please

    Evaluate the limit and indicate appropriate limit laws.

    lim square root of u^4+3u+6
    u->-2

    Evaluate the limit

    lim x^2+5x+4/x^2-3x-4
    x->-4

    Find the limit

    lim t^2+2/t^3+t^2-1
    t->-infinity

    Please can someone help me?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by bballj228 View Post
    Evaluate the limit and indicate appropriate limit laws.

    lim square root of u^4+3u+6
    u->-2

    Evaluate the limit

    lim x^2+5x+4/x^2-3x-4
    x->-4

    Find the limit

    lim t^2+2/t^3+t^2-1
    t->-infinity

    Please can someone help me?
    For the first one all you need to know is this
    Let f(x) be a function defined at some point a, then

    \lim_{x\to{a}}f(x)=f(a)

    For the second one, consider two polynomials

    f(x) and g(x) if

    f(a)=g(a)=0

    Both f(x) and g(x) contain the factor x-a

    Now for your last one Consider multiplying by \frac{\frac{1}{x^2}}{\frac{1}{x^2}}

    from there you can apply the first rule I gave you remembering the lax expression \frac{a}{\infty}=0 where a is a nondescript constant
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  3. #3
    Senior Member nikhil's Avatar
    Joined
    Jun 2008
    Posts
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    Check this out

    Hi!bballj228
    here is the solution
    1))A=Lt [u^4+3u+6]^1/2
    u->-2
    put u =-2 and u will get
    A=[16]^(1/2)
    if u roughly look at the xpression or its graph u will find that the xpression>0 for all values of u.
    therefor A=[16]^(1/2)=+4.
    2)) B=Lt (x^2+5x+4)/(x^2-3x-4) x->-4
    Nox just put x=-4 u will get
    B=0/24=0
    3))C=Lt[t^2+2]/t^3+t^2-1
    t->infinity
    divide numerator and denominator by t^2 u will get
    C=Lt[1+2/(t^2)]/[t+1-1/(t^2)]
    t->infinity
    now put t=infinity
    [remember 1/infinity=0]
    finally u will get
    C=1/(1+infinity) = 0
    (infinity+1=infinity itself)
    hope this helps ya.
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