Use the following to determine the limit. $$ \begin{array}{l}\text { Discuss the end behavior by setting up } \ \text { the limits and evaluating the limits using } \ \qquad \lim _{x \rightarrow \pm \infty} \frac{1}{x^{n}}=\lim _{x \rightarrow \pm \infty} x^{-n}=0 . \ \lim _{x \rightarrow \infty} \frac{1-x^{2}}{x^{3}-x+1}\end{array} $$

Question

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We are given the limit

$$
\lim_{x \rightarrow \infty} \frac{1-x^2}{x^3-x+1}.
$$

**Step 1: Identify the highest powers**

The numerator is a polynomial of degree 2 with leading term $$-x^2$$. The denominator is a polynomial of degree 3 with leading term $$x^3$$.

**Step 2: Divide numerator and denominator by $$x^3$$**

We divide each term by $$x^3$$:

$$
\frac{1-x^2}{x^3-x+1} = \frac{\frac{1}{x^3}-\frac{x^2}{x^3}}{1-\frac{x}{x^3}+\frac{1}{x^3}} = \frac{\frac{1}{x^3}-\frac{1}{x}}{1-\frac{1}{x^2}+\frac{1}{x^3}}.
$$

**Step 3: Analyze the behavior as $$x$$ approaches infinity**

As $$x \rightarrow \infty$$, each term with negative powers of $$x$$ approaches zero:

$$
\frac{1}{x^3} \rightarrow 0,\quad \frac{1}{x} \rightarrow 0,\quad \frac{1}{x^2} \rightarrow 0.
$$

Thus, the expression simplifies to:

$$
\frac{0-0}{1-0+0} = \frac{0}{1} = 0.
$$

**Final Answer:**

$$
\lim_{x \rightarrow \infty} \frac{1-x^2}{x^3-x+1} = 0.
$$
The limit of $$ \frac{1 - x^2}{x^3 - x + 1} $$ as $$ x $$ approaches infinity is 0.

Use the following to determine the limit.

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