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Math Help - limits problem..

  1. #1
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    limits problem..

    Hi guys, could i please get some help on this practice question.

    Consider the function:

     <br />
g(x) = \frac{5x^2}{x^2 + 2x - 3}<br />
for x \notin {-3,1}

    find the following:

    (a) \lim_{x \to +\infty} g(x)

    much appreciated
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  2. #2
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    g(x) = \frac{5x^2}{x^2 + 2x - 3}=5 - \frac{10x-15}{x^2 + 2x - 3}=5 - \frac{\frac{10}{x}-\frac{15}{x^2}}{1+\frac{2}{x}-\frac{3}{x^2}}

    Can you take it from here?
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  3. #3
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    Quote Originally Posted by jvignacio View Post
    Hi guys, could i please get some help on this practice question.

    Consider the function:

     <br />
g(x) = \frac{5x^2}{x^2 + 2x - 3}<br />
for x \notin {-3,1}

    find the following:

    (a) \lim_{x \to +\infty} g(x)

    much appreciated
    "infinity" is hard to deal with but 0 is easy: divide both numerator and denominator by the highest power of x, x^2:
    \frac{5}{1+ \frac{2}{x}- \frac{3}{x^2}}
    Now, as x goes to infinity, the fractions with x in the denominator go to 0.
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  4. #4
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    Quote Originally Posted by Thomas154321 View Post
    g(x) = \frac{5x^2}{x^2 + 2x - 3}=5 - \frac{10x-15}{x^2 + 2x - 3}=5 - \frac{\frac{10}{x}-\frac{15}{x^2}}{1+\frac{2}{x}-\frac{3}{x^2}}

    Can you take it from here?
    sorry im lost .. im not quite sure what you did to get from the first step to the second.
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  5. #5
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    Quote Originally Posted by jvignacio View Post
    sorry im lost .. im not quite sure what you did to get from the first step to the second.
    Algebraic long division. Or just say 5x^2 = (5x^2 + 10x -15) -10x + 15 . Then the numerator can be split up into 5 times the denominator with a remainder term. It's easier to do it like HallsofIvy suggests; I just prefer to take limits that tend to zero where possible with polynomial limits (for me at least it makes the working clearer).
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  6. #6
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    Quote Originally Posted by Thomas154321 View Post
    Algebraic long division. Or just say 5x^2 = (5x^2 + 10x -15) -10x + 15 . Then the numerator can be split up into 5 times the denominator with a remainder term. It's easier to do it like HallsofIvy suggests; I just prefer to take limits that tend to zero where possible with polynomial limits (for me at least it makes the working clearer).
    okay i know how to do Algebraic long division but i still cant seem to understand whats happening
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  7. #7
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    Quote Originally Posted by jvignacio View Post
    okay i know how to do Algebraic long division but i still cant seem to understand whats happening
    What is \lim_{x\rightarrow \infty} \frac{1}{x}?

    What is \lim_{x\rightarrow \infty} \frac{5}{x^2}?

    Do you see the point?
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    What is \lim_{x\rightarrow \infty} \frac{1}{x}?

    What is \lim_{x\rightarrow \infty} \frac{5}{x^2}?

    Do you see the point?
    sorry for not understanding but what point are u trying to show me?
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  9. #9
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    Pay attention and read post #3.
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  10. #10
    Senior Member Pinkk's Avatar
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    The general rule of thumb is to take the limit of the highest power of the numerator over the highest power of the denominator. So in your problem:

    \lim_{x \to +\infty}g(x)=\lim_{x \to +\infty} \frac{5x^{2}}{x^{2}} = \lim_{x \to +\infty} 5 = 5. The above posts are proving to why that is true.

    Quote Originally Posted by jvignacio View Post
    sorry im lost .. im not quite sure what you did to get from the first step to the second.
    He kept dividing until he reached a point where you get:

    \lim_{x \to +\infty} g(x)= 5-\frac{0-0}{1+0-0} =5

    Remember:
    If f(x) = \frac{1}{x^{n}} where n > 0, then:

    \lim_{x \to +\infty} f(x) = 0
    Last edited by Krizalid; March 25th 2009 at 07:19 AM.
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