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    Compute: Lim(x tends to infinity) x^3-x+2/-2x^3+x^2-1 ?
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    (Original post by ersdfrtg)
    Compute: Lim(x tends to infinity) x^3-x+2/-2x^3+x^2-1 ?
    Assuming it is \displaystyle \lim_{x \rightarrow \infty}\frac{x^3-x+2}{-2x^3+x^2-1} then just divide top and bottom by the highest degree of x and observe what happens to each term with x tending to infinity before evaluating what happens with the overall fraction.
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    (Original post by RDKGames)
    Assuming it is \displaystyle \lim_{x \rightarrow \infty}\frac{x^3-x+2}{-2x^3+x^2-1} then just divide top and bottom by the highest degree of x and observe what happens to each term with x tending to infinity before evaluating what happens with the overall fraction.
    And how would you divide this function by the highest degree of x? I tried doing long division, but it didn't work for me :/
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    (Original post by ersdfrtg)
    And how would you divide this function by the highest degree of x? I tried doing long division, but it didn't work for me :/
    The highest degree of x in this expression is 3 so divide top and bottom of this fraction by x^3.
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    (Original post by notnek)
    The highest degree of x in this expression is 3 so divide top and bottom of this fraction by x^3.
    But then that will give me negative powers: 1-x^-2+2x^-3 / -2+x^-2-x^-3
    ?
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    (Original post by ersdfrtg)
    But then that will give me negative powers: 1-x^-2+2x^-3 / -2+x^-2-x^-3
    ?
    It should be x^-1 instead of x^-2 on the bottom but other than that it's correct. Then the fraction can be written as

    \displaystyle \frac{1-\frac{1}{x^2} + \frac{2}{x^3}}{-2+\frac{1}{x^2} - \frac{1}{x^3}}

    Go through each term in this fraction and think about what happens as x\rightarrow \infty

    e.g. As x\rightarrow \infty, \frac{1}{x^2} tends to 0.
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    x^3-x+2/-2x^3+x^2-1 ---> inf as x ---> inf

    Do they not teach brackets anymore?
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    (Original post by mik1a)
    x^3-x+2/-2x^3+x^2-1 ---> inf as x ---> inf

    No it doesn't.
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    (Original post by RDKGames)
    No it doesn't.
    It does if you interpret brackets correctly. The x^3 is the dominant term.
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    (Original post by mik1a)
    It does if you interpret brackets correctly. The x^3 is the dominant term.
    Are you interpreting it as \displaystyle \lim_{x\rightarrow \infty} (x^3-x+\frac{2}{-2x^3}+x^2-1)? In which case that would be correct, otherwise I do not know what brackets you're talking about.
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    Yes, hence "do they not teach brackets anymore?"
 
 
 
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