find the lim as x approaches positive and negative infinity
(3x^3 - x +1) / (x + 3)
can you explain the answer
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 $\displaystyle \displaystyle \begin{align*} -\infty \end{align*}$...
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.