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Math Help - Horizontal Asymptote

  1. #1
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    Horizontal Asymptote

    Ok, we learned limits a while back so I am trying to scrape away the cobwebs, I am having trouble finding the horizontal asymptote to a function to which I eventually will need to sketch a curve. My professor said to use the power rule just like we used to evaluate limits, by multiplying by the reciprocal, so I multiplied each term by (1/X) but I cant get the value the book has for the asymptote. Here is the problem:

    F(X)= 3X-1 / X+2

    Book has answer of Y=3 for horizontal, can someone illustrate how I use the reciprocal power rule for this, thanks!!
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  2. #2
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    Something to take note of is when you are taking the lmit of a rational expression, everything but the highest powers can be disregarded.

    This leaves you with \frac{3x}{x}=3


    Also, just divide and you can see it right off.


    \frac{3x-1}{x+2}=3-\frac{7}{x+2}

    \lim_{x\rightarrow{\infty}}\left[3-\frac{7}{x+2}\right]=?


    What your professor is probably talking about is dividing the top and bottom by x.

    \lim_{x\rightarrow{\infty}}\frac{\frac{3x}{x}-\frac{2}{x}}{\frac{x}{x}+\frac{2}{x}}=

    \lim_{x\rightarrow{\infty}}\frac{3-\frac{1}{x}}{1+\frac{2}{x}}

    As you can see, the terms with the x in the denominator tend to 0 and you are left with 3.
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  3. #3
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    What about this one:

    X^2 + 2 / X^2+1

    I used the reciprocal of 1/X^2 and got

    1+(2/X^2) / 1+X^2 but I don't find where this yields the asymptote of Y=1 as is the answer. What am I doing wrong?
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  4. #4
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    You are correct, it's right in front of you.

    \frac{\frac{x^{2}}{x^{2}}+\frac{2}{x^{2}}}{\frac{x  ^{2}}{x^{2}}+\frac{1}{x^{2}}}

    \frac{1+\frac{2}{x^{2}}}{1+\frac{1}{x^{2}}}

    Same as before, the terms with x^2 in the denominator tend to 0 as x-->infinity and you are left with 1 as the limit.
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  5. #5
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    So looking at both examples it looks like I just need to recognize that whatever constant term I get will be my limit? It seems like this is almost too easy. This looks totally different than the way we did it in class, but I guess if it works. Are we still following the reciprocal power rules method to obtain the asymptote?
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  6. #6
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    rational expressions are among the easiest to find limits.

    Disregard everything but the terms with the highest powers.

    If the denominator has the highest terms, then its limit is 0.

    If the powers are the same, then the limit is the ratio of the leading coefficients.

    If the numerator has the highest power, then the limit is infinity.
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  7. #7
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    Quote Originally Posted by galactus View Post
    rational expressions are among the easiest to find limits.

    Disregard everything but the terms with the highest powers.

    If the denominator has the highest terms, then its limit is 0.

    If the powers are the same, then the limit is the ratio of the leading coefficients.

    If the numerator has the highest power, then the limit is infinity.
    I am not looking for limits though, just the asymptote, but your saying just look at the term in the function that has the highest power and divide the coefficients and that will be my asymptote?
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  8. #8
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    That is how you find a horizontal asymptote. By taking

    \lim_{x\rightarrow{\infty}}f(x)=L

    or

    \lim_{x\rightarrow{-\infty}}f(x)=L

    That's what an asymptote is.
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