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Math Help - square roots and negative infinity

  1. #1
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    square roots and negative infinity

    problem:
    Find the limit. (If you need to use - or , enter -INFINITY or INFINITY.)


    I would explain what ive done in more detail, but i know im missing something due to x approaching neg. infinity, but i did multiply the above funct. by its conjugate to get (x-1)/(x-square root of x^2+4x). How would i go about this? Thank you. -Evan
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    Quote Originally Posted by Evan.Kimia View Post
    problem:
    Find the limit. (If you need to use - or , enter -INFINITY or INFINITY.)


    I would explain what ive done in more detail, but i know im missing something due to x approaching neg. infinity, but i did multiply the above funct. by its conjugate to get (x-1)/(x-square root of x^2+4x). How would i go about this? Thank you. -Evan
    Let x = -t and calculate \lim_{t \rightarrow + \infty} \left( \sqrt{ t^2 - 4t} - t\right).
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    Ok here is what i have...

    sqroot(t^2-4t)-t * sqroot(t^2-4t)+t
    ------------------
    1 * sqroot(t^2-4t)+t

    t^2-4t-t^2
    -------------
    sqroot(t^2-4t)+t

    consolidate like terms then divide by t....
    -4
    ---------------
    * sqroot(t^2-4t)+1


    how can i properly solve for the denominator with the sqroot involved? I have something on paper that gets me sqroot(t-4)+1 which is most likely wrong. Thanks.
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    Quote Originally Posted by Evan.Kimia View Post
    Ok here is what i have...

    sqroot(t^2-4t)-t * sqroot(t^2-4t)+t
    ------------------
    1 * sqroot(t^2-4t)+t

    t^2-4t-t^2
    -------------
    sqroot(t^2-4t)+t

    consolidate like terms then divide by t....
    -4
    ---------------
    * sqroot(t^2-4t)+1


    how can i properly solve for the denominator with the sqroot involved? I have something on paper that gets me sqroot(t-4)+1 which is most likely wrong. Thanks.
    When you divided the denominator by t, you failed to divide the square root term by t. The correct expression is \frac{-4}{\sqrt{1 - \frac{4}{t}} + 1}. Now take the required limit.
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    I see. How would i properly divide t by the denominator? I know this is easy, but my algebra is off. I know dividing by t in the sqroot wont really work unless t itself is within a square root, so t= sqroot(t^2), am i right? If thats so, then i can see sqroot(t^2-4t)
    turning into (1-4/t), but wouldnt the 1 turn into 1/sqroot(x^2) which would equal 0? I know im wrong, just thinking out loud.

    Also, you explained this better than my math teacher today, thanks for everything.
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    Quote Originally Posted by Evan.Kimia View Post
    I see. How would i properly divide t by the denominator? I know this is easy, but my algebra is off. I know dividing by t in the sqroot wont really work unless t itself is within a square root, so t= sqroot(t^2), am i right? If thats so, then i can see sqroot(t^2-4t)
    turning into (1-4/t), but wouldnt the 1 turn into 1/sqroot(x^2) which would equal 0? I know im wrong, just thinking out loud.

    Also, you explained this better than my math teacher today, thanks for everything.
    You divide each term in the denominator by t.

    Dividing the square root term by t works in the way you have described. And when you divide the other term by t (and that other term is t) then you get t/t = 1. So you end up with a denominator as I posted.
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