# Thread: Vertical asymptote and limit (left and right)

1. ## 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.
$
y = \frac{{1 + x^4 }}{{x^2 - x^4 }}
$

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

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

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
$
y=f(x)=\frac{{1 + x^4 }}{{x^2 - x^4 }}
$
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
$
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.

$
\begin{array}{l}
\mathop {\lim }\limits_{x \to 0^ - } = ? \\
\mathop {\lim }\limits_{x \to 0^ + } = ? \\
\mathop {\lim }\limits_{x \to 1^ - } = ? \\
\mathop {\lim }\limits_{x \to 1^ + } = ? \\
\mathop {\lim }\limits_{x \to - 1^ - } = ? \\
\mathop {\lim }\limits_{x \to - 1^ + } = ? \\
\end{array}
$

2. Originally Posted by Mathlv
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.
$
y = \frac{{1 + x^4 }}{{x^2 - x^4 }}
$

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

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

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
$
y=f(x)=\frac{{1 + x^4 }}{{x^2 - x^4 }}
$
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
$
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.

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

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

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 $\color{blue}+\infty$

3. ## Vertical asymptote

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