Resolved: "Find limx→−∞3x3−2xx2+2x+8"

geduiwelh

Answered question

2021-10-15

Find

lim_{x \to - \infty} \frac{3 x^{3} - 2 x}{x^{2} + 2 x + 8}

Answer & Explanation

pivonie8

Skilled2021-10-16Added 91 answers

In order to find the value of $lim_{x \to - \infty} \frac{3 x^{3} - 2 x}{x^{2} + 2 x + 8}$ , first divide the numerator and the denominator by the highest power in the denominator, that is, $x^{2}$
$lim_{x \to - \infty} \frac{3 (\frac{x^{3}}{x^{2}}) - 2 \frac{x}{x^{2}}}{\frac{x^{2}}{x^{2}} + 2 \frac{x}{x^{2}} + 8 \frac{1}{x^{2}}} = lim_{x \to - \infty} \frac{3 x - 2 \frac{1}{x}}{1 + 2 \frac{1}{x} + 8 \frac{1}{x^{2}}}$
Substitute $- \infty$ insteaf of x in the numerator and the denomiantor of the above fraction.
$lim_{x \to - \infty} \frac{3 x - \frac{2}{x}}{1 + \frac{2}{x} + \frac{8}{x^{2}}} = \frac{3 (- \infty) - \frac{2}{(- \infty)}}{1 + \frac{2}{- \infty} + \frac{8}{{(- \infty)}^{2}}}$
$= \frac{- \infty - 0}{1 + 0 + 0}$
$= \frac{- \infty}{1}$
$= - \infty$
Answer: The value of $lim_{x \to - \infty} \frac{3 x^{3} - 2 x}{x^{2} + 2 x + 8}$ is $- \infty$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?