Question

math 220 - written homework 9
due: wed october 29 at 6:00 pm
4.6 limits at infinity, asymptotes

1. evaluate the limits at infinity

$$\lim_{x \to -\infty} \frac{3x^{5}+x^{4}-7x + 1}{2x^{4}+x^{3}-x + 1}$$
$$\lim_{x \to -\infty} \frac{3x^{4}+x^{4}-7x + 1}{x - x^{2}}$$
$$\lim_{x \to \infty} \frac{3x^{3}+x^{2}-7x + 1}{2x^{4}-3x + 5}$$
$$\lim_{x \to -\infty} \sqrt{x^{2}+2x - 3}$$

Explanation:

First Limit: \(\lim_{x

ightarrow-\infty}\frac{3x^{5}+x^{4}-7x + 1}{2x^{4}+x^{3}-x + 1}\)

Step1: Divide numerator and denominator by \(x^{4}\)

$$
LATEXBLOCK0
$$

Step2: Evaluate the limit

As \(x
ightarrow-\infty\), \(\frac{1}{x}
ightarrow0\), \(\frac{1}{x^{3}}
ightarrow0\), \(\frac{1}{x^{4}}
ightarrow0\). So, \(\lim_{x
ightarrow-\infty}\frac{3x + 1-\frac{7}{x^{3}}+\frac{1}{x^{4}}}{2+\frac{1}{x}-\frac{1}{x^{3}}+\frac{1}{x^{4}}}=-\infty\)

Second Limit: \(\lim_{x

ightarrow-\infty}\frac{3x^{4}+x^{4}-7x + 1}{x - x^{2}}\)

Step1: Simplify the numerator and divide by \(x^{2}\)

The numerator \(3x^{4}+x^{4}-7x + 1 = 4x^{4}-7x + 1\). Divide numerator and denominator by \(x^{2}\):

$$
LATEXBLOCK0
$$

Step2: Evaluate the limit

As \(x
ightarrow-\infty\), \(\frac{1}{x}
ightarrow0\), \(\frac{1}{x^{2}}
ightarrow0\). So, \(\lim_{x
ightarrow-\infty}\frac{4x^{2}-\frac{7}{x}+\frac{1}{x^{2}}}{\frac{1}{x}-1}=-\infty\)

Third Limit: \(\lim_{x

ightarrow\infty}\frac{3x^{3}+x^{2}-7x + 1}{2x^{4}-3x + 5}\)

Step1: Divide numerator and denominator by \(x^{4}\)

$$
LATEXBLOCK0
$$

Step2: Evaluate the limit

As \(x
ightarrow\infty\), \(\frac{1}{x}
ightarrow0\), \(\frac{1}{x^{2}}
ightarrow0\), \(\frac{1}{x^{3}}
ightarrow0\), \(\frac{1}{x^{4}}
ightarrow0\). So, \(\lim_{x
ightarrow\infty}\frac{\frac{3}{x}+\frac{1}{x^{2}}-\frac{7}{x^{3}}+\frac{1}{x^{4}}}{2-\frac{3}{x^{3}}+\frac{5}{x^{4}}}=0\)

Fourth Limit: \(\lim_{x

ightarrow-\infty}\sqrt{x^{2}+2x - 3}\)

Answer:

1. \(-\infty\)
2. \(-\infty\)
3. \(0\)
4. \(\infty\)