Question

Limit as x-> neg. infinity of
[(-3x^3+4x+1) / (2-x(x^2-1))]

Answer 1

\[\lim_{x \rightarrow -\infty} [ -3x ^{3}+4x +1) / (2-x(x ^{2}-1) )]\]

Answer 2

\[\lim_{x \rightarrow -\infty} \frac{-3x^{3}+4x+1}{2-x(x^{2}-1)}\]?

Answer 3

yes

Answer 4

\[\lim_{x \rightarrow -\infty} \frac{-3x^{3}+4x+1}{2-x^{3}-x} \]\[\lim_{x \rightarrow -\infty} \frac{-3x^{3}+4x+1}{2-x^{3}-x} \times \left(\frac{\frac{1}{x^{3}}}{\frac{1}{x^{3}}} \right)\]\[\lim_{x \rightarrow -\infty} \frac{-3+\frac{4}{x^{2}}+\frac{1}{x^{3}}}{ \frac{2}{x^{3}}-1-\frac{1}{x^{2}} } \]Evaluate at -infinity:\[ \frac{-3+\frac{4}{(-\infty)^{2}}+\frac{1}{(-\infty)^{3}}}{ \frac{2}{(-\infty)^{3}}-1-\frac{1}{(-\infty)^{2}} } \] \[ \frac{-3+0+0}{ 0-1-0 } = 3\]
http://www.wolframalpha.com/input/?i=limit+as+x+goes+to+-infinity+of+%28-3x^3%2B4x%2B1%29%2F%282-x%28x^2-1%29%29

Answer 5

THANK YOU SOO MUCH !!!!

Answer 6

btw wat is teh L-hospital rule...i see taht alot but my teacher nvr taught us anythg with taht