find the lim as x approaches positive and negative infinity

(3x^3 - x +1) / (x + 3)

can you explain the answer

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- Sep 29th 2012, 07:04 AMubhuttoNeed LImit help!!!!!!!!!!
find the lim as x approaches positive and negative infinity

(3x^3 - x +1) / (x + 3)

can you explain the answer - Sep 29th 2012, 07:07 AMProve ItRe: Need LImit help!!!!!!!!!!
Start by dividing the top and bottom by x, to give

$\displaystyle \displaystyle \begin{align*} \frac{3x^2 - 1 + \frac{1}{x}}{1 + \frac{3}{x}} \end{align*}$

Then as you approach $\displaystyle \displaystyle \begin{align*} \infty \end{align*}$ the top goes to $\displaystyle \displaystyle \begin{align*} \infty \end{align*}$ and the bottom goes to 1, so the limit is $\displaystyle \displaystyle \begin{align*} \infty \end{align*}$.

Try evaluating the limit as you approach $\displaystyle \displaystyle \begin{align*} -\infty \end{align*}$... - Sep 29th 2012, 07:11 AMubhuttoRe: Need LImit help!!!!!!!!!!
thanks

can you also show me how to get the horizontal asymptote of the graph - Sep 29th 2012, 11:23 AMHallsofIvyRe: Need LImit help!!!!!!!!!!
Another way to do this to go ahead and divide: $\displaystyle \frac{3x^3- x+ 1}{x+1}= 3x^2- 3x+ 2- \frac{1}{x+1}$. As x goes to either + or - infinity, that last fraction goes to 0 so you only need to consider what happens to $\displaystyle 3x^2- 3x+ 2$ determines the limit.

If you know what a "horizontal asymptote" is, you should realize that knowing the limit tells you everything you need to know about the horizontal asymptotes.