Question

hw 4 - limits to infinity section 1.3: problem 8 (1 point) find the following limit. notes: enter \dne\ if the the limit does not exist. \\( \lim_{x \to -\infty} \frac{6x - 2x^3 + 4}{5x^3 + x^2 + 1} = \square \\)

Explanation:

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

For the numerator \(6x - 2x^3 + 4\), dividing each term by \(x^3\) gives \(\frac{6x}{x^3}-\frac{2x^3}{x^3}+\frac{4}{x^3}=\frac{6}{x^2}-2 + \frac{4}{x^3}\).
For the denominator \(5x^3 + x^2 + 1\), dividing each term by \(x^3\) gives \(\frac{5x^3}{x^3}+\frac{x^2}{x^3}+\frac{1}{x^3}=5+\frac{1}{x}+\frac{1}{x^3}\).

So the limit becomes \(\lim_{x
ightarrow -\infty}\frac{\frac{6}{x^2}-2 + \frac{4}{x^3}}{5+\frac{1}{x}+\frac{1}{x^3}}\).

Step2: Evaluate limits of each term

As \(x
ightarrow -\infty\), \(\frac{6}{x^2}
ightarrow 0\) (since \(x^2\) is positive and large, reciprocal is small), \(\frac{4}{x^3}
ightarrow 0\) (since \(x^3\) is negative and large in magnitude, reciprocal is small), and \(\frac{1}{x}
ightarrow 0\) (since \(x\) is negative and large in magnitude, reciprocal is small).

Substituting these into the expression, we get \(\frac{0 - 2+0}{5 + 0+0}\).

Step3: Simplify the fraction

Simplify \(\frac{-2}{5}\).

Answer:

\(-\frac{2}{5}\)