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.

$\displaystyle

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:

$\displaystyle

\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

$\displaystyle

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 $\displaystyle + \infty $ or $\displaystyle -\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.

$\displaystyle

\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}

$