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Math Help - evaluating limits

  1. #1
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    evaluating limits

    Hi

    I know I need to factor these but I just can't seem to figure out how

    First


    lim (x+3)/(√(9x^2-5x))
    x-> infinity

    Second

    lim ((√(x^2+1))-x)
    x ->infinity


    Thanks in advance
    from this calculus beginner
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  2. #2
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    Quote Originally Posted by calcbeg View Post
    Hi

    I know I need to factor these but I just can't seem to figure out how

    First


    lim (x+3)/(√(9x^2-5x))
    x-> infinity

    Second

    lim ((√(x^2+1))-x)
    x ->infinity


    Thanks in advance
    from this calculus beginner

    lim (x+3)/(√(9x^2-5x))
    x-> infinity

    This limit can be reduced to this


    lim (x+3)/(|x|√(9-5/x))
    x-> infinity

    All you have to remember is that √x^2 = |x|

    So if you factor out x^2 from under the square root you get

    lim (x+3)/(√(x^2(9-5/x))
    x-> infinity

    Which is the same thing as

    lim (x+3)/(|x|√(9-5/x))
    x-> infinity

    x --> infinity so x is always +


    Divide everything by x
    lim (1+3/x )/(√(9-5/x)) =1/3
    x-> infinity

    2)

    lim ((√(x^2+1))-x)
    x ->infinity

    divide everything by x like the first question

    lim ((√(1+1/x^2))-1)/(1/x)
    x ->infinity

    use a substitution t = 1/x

    lim ((√(1+t^2)-1)/(t)
    t->0

    You can see that this is really the definition of a derivative just with t not h.

    lim [f(x+h) - f(h)]/ h
    h --> 0

    f(x) = √(1+t^2)

    f '(x) = 2t /2(√(1+t^2)

    f ' (0) = 0 /2 = 0

    So the limit = 0
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  3. #3
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    Joined
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    Quote Originally Posted by calcbeg View Post
    Hi

    I know I need to factor these but I just can't seem to figure out how

    First


    lim (x+3)/(√(9x^2-5x))
    x-> infinity

    Second

    lim ((√(x^2+1))-x)
    x ->infinity


    Thanks in advance
    from this calculus beginner


    1) If x>5/9..... \lim_{x\to\infty}{\frac{\ x+3}{\sqrt{9x^2-5x}}}=\lim_{x\to\infty}{\frac{\ 1+\frac{\ 3}{x}}{\sqrt{9-\frac{\ 5}{x}}}}=\frac{\ 1+3.0}{\sqrt{9-5.0}}=\frac{\ 1}{3}.


    2) If now we multiply \sqrt{x^2+1} -x by \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x},we get:


    \frac{\ 1}{\sqrt{x^2+1}+x} and if.......x>0..... \lim_{x\to\infty}{\frac{\ 1}{\sqrt{x^2+1}+x}}=\lim_{x\to\infty}{\frac{\frac{  \ 1}{x}}{\sqrt{1+\frac{\ 1}{x^2}}+1}}=\frac{\ 0}{\sqrt{1+0}+1}=0
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