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Math Help - Vertical asymptote and limit (left and right)

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    33

    Vertical asymptote and limit (left and right)

    Hello,
    First let me describe the problem and what I have done so far:
    The Problem:

    Find the horizontal and vertical asymptotes of the curve. Estimating the asymptotes.
    <br />
y = \frac{{1 + x^4 }}{{x^2  - x^4 }}<br />

    For the horizontal asymptote, I was able to find it, and it is -1.

    For the vertical asymptote, I find it as follow:
    <br />
\begin{array}{l}<br />
 x^2  - x^4  = 0 \Rightarrow x^2 \left( {1 - x^2 } \right) = 0 \\ <br />
 x^2 \left( {1 + x} \right)\left( {1 - x} \right) = 0 \\ <br />
 x = 0 \\ <br />
 1 + x = 0 \Rightarrow x =  - 1 \\ <br />
 1 - x = 0 \Rightarrow x = 1 \\ <br />
 \end{array}<br />

    So when x=0, x=1, & x=-1 they will cause the denominator = 0, but the numerator is not = 0; therefore, I have to investigate
    <br />
y=f(x)=\frac{{1 + x^4 }}{{x^2  - x^4 }}<br />
as x approaches these points x=0, x=1, & x=-1 from the left and from the right of these points. THIS IS WHERE I AM HAVING A PROBLEM.
    QUESTION: Would you please explain to me how I can find these values: I am guessing that the values possibly either  + \infty or  -\infty <br />
but I do not know how to tell which one is the correct one unless I create a table and graph it. I believe there must be a way to tell it without having to graph it out.

    <br />
\begin{array}{l}<br />
 \mathop {\lim }\limits_{x \to 0^ -  }  = ? \\ <br />
 \mathop {\lim }\limits_{x \to 0^ +  }  = ? \\ <br />
 \mathop {\lim }\limits_{x \to 1^ -  }  = ? \\ <br />
 \mathop {\lim }\limits_{x \to 1^ +  }  = ? \\ <br />
 \mathop {\lim }\limits_{x \to  - 1^ -  }  = ? \\ <br />
 \mathop {\lim }\limits_{x \to  - 1^ +  }  = ? \\ <br />
 \end{array}<br />
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  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Mathlv View Post
    Hello,
    First let me describe the problem and what I have done so far:
    The Problem:

    Find the horizontal and vertical asymptotes of the curve. Estimating the asymptotes.
    <br />
y = \frac{{1 + x^4 }}{{x^2  - x^4 }}<br />

    For the horizontal asymptote, I was able to find it, and it is -1.

    For the vertical asymptote, I find it as follow:
    <br />
\begin{array}{l}<br />
 x^2  - x^4  = 0 \Rightarrow x^2 \left( {1 - x^2 } \right) = 0 \\ <br />
 x^2 \left( {1 + x} \right)\left( {1 - x} \right) = 0 \\ <br />
 x = 0 \\ <br />
 1 + x = 0 \Rightarrow x =  - 1 \\ <br />
 1 - x = 0 \Rightarrow x = 1 \\ <br />
 \end{array}<br />

    So when x=0, x=1, & x=-1 they will cause the denominator = 0, but the numerator is not = 0; therefore, I have to investigate
    <br />
y=f(x)=\frac{{1 + x^4 }}{{x^2  - x^4 }}<br />
as x approaches these points x=0, x=1, & x=-1 from the left and from the right of these points. THIS IS WHERE I AM HAVING A PROBLEM.
    QUESTION: Would you please explain to me how I can find these values: I am guessing that the values possibly either  + \infty or  -\infty <br />
but I do not know how to tell which one is the correct one unless I create a table and graph it. I believe there must be a way to tell it without having to graph it out.

    <br />
\begin{array}{l}<br />
 \mathop {\lim }\limits_{x \to 0^ -  }  = ? \\  <br />
 \mathop {\lim }\limits_{x \to 0^ +  }  = ? \\ <br />
 \mathop {\lim }\limits_{x \to 1^ -  }  = ? \\ <br />
 \mathop {\lim }\limits_{x \to 1^ +  }  = ? \\ <br />
 \mathop {\lim }\limits_{x \to  - 1^ -  }  = ? \\ <br />
 \mathop {\lim }\limits_{x \to  - 1^ +  }  = ? \\ <br />
 \end{array}<br />
    \mathop {\lim }\limits_{x \to 0^ -  }\frac{1+x^4}{x^2(1-x)(1+x)}

    1+x^4>0,\ x^2>0,\ 1+x>0,\ 1-x>0

    graph goes to \color{blue}+\infty

    \mathop {\lim }\limits_{x \to 0^ +  }\frac{1+x^4}{x^2(1-x)(1+x)}

    1+x^4>0,\ x^2>0,\ 1+x>0,\ 1-x>0

    graph goes to \color{blue}+\infty

    \mathop {\lim }\limits_{x \to 1^ -  }\frac{1+x^4}{x^2(1-x)(1+x)}

    1+x^4>0,\ x^2>0,\ 1+x>0,\ 1-x>0

    graph goes to \color{blue}+\infty

    \mathop {\lim }\limits_{x \to 1^ +  }\frac{1+x^4}{x^2(1-x)(1+x)}

    1+x^4>0,\ x^2>0,\ 1+x>0,\ \color{red}1-x<0

    graph goes to \color{red}-\infty

    \mathop {\lim }\limits_{x \to  - 1^ -  }\frac{1+x^4}{x^2(1-x)(1+x)}

    1+x^4>0,\ x^2>0,\ \color{red}1+x<0,\ \color{black}1-x>0

    graph goes to \color{red}-\infty

    \mathop {\lim }\limits_{x \to  - 1^ +  }\frac{1+x^4}{x^2(1-x)(1+x)}

    1+x^4>0,\ x^2>0,\ 1+x>0,\ 1-x>0

    graph goes to \color{blue}+\infty
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  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    33

    Vertical asymptote

    Thank you very much for the help and explanation Archie Meade.
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