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Math Help - Help With Limits!

  1. #1
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    Help With Limits!

    Question#1.Determine the Values for c and N so that:

    lim
    x--->infinity

    <br />
\frac {10*x^5-15*x^3+2}{c*x^n-6*x^2+18}<br />

    And the above equation equals ( I suck at this latex stuff)

    <br />
\frac {10}{7}<br />

    Which equals:1.42857142857143

    Solve for C and N


    Question#2.
    Let f(x)=9*x^8-12*x^3+3 and g(x)=cx^n-6*x^2+18 with c cant be 0. Then :

    Lim
    x---->infinity

    <br />
\frac {fx}{gx}<br />

    Equals to 0

    This Implies that N is either less than, equal to, or more than ______?

    I know my latex is pretty bad so if anything looks confusing, let me know and I'll try to fix it.

    I think I know the first one. I believe the value for C must be 7 and N must be 4 since there is a horizontal assymtote. Correct? On the right path?

    I am lost on the second one.
    Last edited by mvho; May 26th 2009 at 09:01 PM.
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  2. #2
    Senior Member
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    Hello!

    Quote Originally Posted by mvho View Post
    Question#1.Determine the Values for c and N so that:

    lim
    x--->infinity

    <br />
\frac {10*x^5-15*x^3+2}{c*x^n-6*x^2+18}<br />

    And the above equation equals ( I suck at this latex stuff)

    <br />
\frac {10}{7}<br />

    Which equals:1.42857142857143

    Solve for C and N
    The solution should be 10/7, it's constant, therefor n must be equal to 5 => n = 5

    Then we have

    <br />
\frac {10*x^5-15*x^3+2}{c*x^5-6*x^2+18}<br />

    This is equal to

    <br />
 \frac {10*x^5}{c*x^5-6*x^2+18}-\frac{15*x^3}{c*x^5-6*x^2+18}+\frac{2}{c*x^5-6*x^2+18}<br />

    Now consider

    lim_{x \to \infty} \ [ \frac {10*x^5}{c*x^5-6*x^2+18}-\frac{15*x^3}{c*x^5-6*x^2+18}+\frac{2}{c*x^5-6*x^2+18} ]

    = lim_{x \to \infty} \ \frac {10*x^5}{c*x^5-6*x^2+18}-lim_{x \to \infty} \  \frac{15*x^3}{c*x^5-6*x^2+18}

    + lim_{x \to \infty} \  \frac{2}{c*x^5-6*x^2+18} ]

    = lim_{x \to \infty} \ \frac {10*x^5}{c*x^5-6*x^2+18} - 0 + 0

    In this case you only have to consider the x-terms with the biggest exponent (here it is x^5, this will tell you the limes)

    Therefor the same problem is

    = lim_{x \to \infty} \ \frac {10*x^5}{c*x^5-6*x^2+18}

    = lim_{x \to \infty} \ \frac {10*x^5}{c*x^5}

    and the solution should be 10/7, I guess you will be able to find c = 7
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  3. #3
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    Quote Originally Posted by mvho View Post
    Question#2.
    Let f(x)=9*x^8-12*x^3+3 and g(x)=cx^n-6*x^2+18 with c cant be 0. Then :

    Lim
    x---->infinity

    <br />
\frac {fx}{gx}<br />

    Equals to 0

    This Implies that N is either less than, equal to, or more than ______?

    ...

    I am lost on the second one.
    In these cases it is only important to check on the x-term with the highest exponents,

    for f it is 8

    for g it is n.

    Let's determine three cases

    if n=8:
    f/g -> const; x to infty

    if n < 8
    f/ g -> infty, x to infty

    if n > 8

    f/g -> 0; x to infty

    because you have something like this

    \frac{12x^8}{0.5x^9} = \frac{12}{0.5x} \to \infty, \ x\to \infty

    Did you understand?

    Yours
    Rapha
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  4. #4
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    Thanks Man, Looks like I made a mistake and N should be 5 instead of 4.
    I never really understood the logic since I just know the short cut that for

    Horizontal assymtotes, you will divide the coefficient in front of the numerator by the denonminator if the degree is greater on the top. If the degrees are the same, then it will just approach 0.
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  5. #5
    Senior Member
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    Quote Originally Posted by mvho View Post
    Thanks Man, Looks like I made a mistake and N should be 5 instead of 4.
    You're welcome.

    Quote Originally Posted by mvho View Post
    I never really understood the logic since I just know the short cut that for

    Horizontal assymtotes, you will divide the coefficient in front of the numerator by the denonminator if the degree is greater on the top.
    Not sure, what you are talking about.
    If the degree of of the numerator f(x) is greater than the degree of the denominator g(x), then \lim_{x \to \infty} \frac{f(x)}{g(x)} = \infty

    If the degree of of the numerator f(x) is smaller than the degree of the denominator g(x), then \lim_{x \to \infty} \frac{f(x)}{g(x)} = 0


    Quote Originally Posted by mvho View Post
    If the degrees are the same, then it will just approach 0.
    No, probably not. It approaches a constant in \mathbb{R}
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  6. #6
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    In general, with a rational function, \frac{a_n x^n + a_{n-1} x^{n-1} + \cdot \cdot \cdot + a_1 x + a_0}{b_m x^m + b_{m-1} x^{m-1} + \cdot \cdot \cdot + b_1 x+ b_0}, as x goes to infinity, you can divide both numerator and denominator by x^m and get \frac{a_nx^{n-m}+ a_{n-1}x^{n-m-1}+ \cdot\cdot\cdot+ a_1x^{1-m}+ a_0x^{-m}}{b_m+ b_{m-1}x^{-1}+ \cdot\cdot\cdot+ b_1x^{1-m}+ b_2x^{-m}}.

    Now, as x goes to infinity, all negative powers of x go to 0 so the denominator goes to b_m. What happens in the numerator depends on the relation between m and n. If n> m, x^{n-m} is still a positive power so the numerator goes to infinity and, since the denominator is going to a finite, non-zero power, the entire fraction goes to infinity. If n< m, then every power of x in the numerator is negative so the numerator goes to 0. Since the denominator goes to a finite, non-zero number, the fraction goes to 0. Finally, if n= m, all terms in the numerator except the first have negative power and go to 0 so the numerator goes to a_n and the fraction goes to \frac{a_n}{b_m}= \frac{a_n}{b_n} since n= m.
    Last edited by mr fantastic; May 27th 2009 at 06:32 AM. Reason: Fixed latex
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  7. #7
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    Thanks everyone, question solved!
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