Question

find the limit. (if the limit is infinite, enter ∞ or -∞, as appropriate. if the limit does not otherwise exist, enter dne.)
\lim_{x \to \infty} \frac{x^4 - 2x^2 + x}{x^3 - x + 4}

Explanation:

Step1: Divide numerator and denominator by \(x^3\)

To find the limit as \(x\to\infty\) of a rational function, we can divide each term in the numerator and the denominator by the highest power of \(x\) in the denominator. The highest power of \(x\) in the denominator \(x^3 - x + 4\) is \(x^3\). So we divide each term in the numerator \(x^4 - 2x^2 + x\) and the denominator \(x^3 - x + 4\) by \(x^3\).

For the numerator:

• \(\frac{x^4}{x^3}=x\)
• \(\frac{-2x^2}{x^3}=-\frac{2}{x}\)
• \(\frac{x}{x^3}=\frac{1}{x^2}\)

For the denominator:

• \(\frac{x^3}{x^3}=1\)
• \(\frac{-x}{x^3}=-\frac{1}{x^2}\)
• \(\frac{4}{x^3}=\frac{4}{x^3}\)

So the function becomes:
\[
\lim_{x\to\infty}\frac{x - \frac{2}{x}+\frac{1}{x^2}}{1-\frac{1}{x^2}+\frac{4}{x^3}}
\]

Step2: Evaluate the limit as \(x\to\infty\)

As \(x\to\infty\), the terms with \(x\) in the denominator will approach \(0\). That is:

• \(\lim_{x\to\infty}\frac{2}{x} = 0\) (because as \(x\) becomes very large, \(\frac{2}{x}\) becomes very small, approaching \(0\))
• \(\lim_{x\to\infty}\frac{1}{x^2}=0\) (similarly, as \(x\) is large, \(x^2\) is very large, so \(\frac{1}{x^2}\) approaches \(0\))
• \(\lim_{x\to\infty}\frac{1}{x^2}=0\) (same reasoning as above)
• \(\lim_{x\to\infty}\frac{4}{x^3}=0\) (as \(x\) is large, \(x^3\) is very large, so \(\frac{4}{x^3}\) approaches \(0\))

And \(\lim_{x\to\infty}x=\infty\) (as \(x\) approaches infinity, \(x\) itself approaches infinity).

Substituting these limits into the function \(\frac{x - \frac{2}{x}+\frac{1}{x^2}}{1-\frac{1}{x^2}+\frac{4}{x^3}}\), we get:

\[
\frac{\lim_{x\to\infty}x-\lim_{x\to\infty}\frac{2}{x}+\lim_{x\to\infty}\frac{1}{x^2}}{\lim_{x\to\infty}1-\lim_{x\to\infty}\frac{1}{x^2}+\lim_{x\to\infty}\frac{4}{x^3}}=\frac{\infty - 0 + 0}{1 - 0+0}=\infty
\]

Answer:

\(\infty\)