Lim (3x^3 +3x^2) / (x - 6x^2)
as x approaches negative infinity

is the answer positive infinity, if not then can you explain

2. ## Re: Limits help please

Originally Posted by ubhutto
Lim (3x^3 +3x^2) / (x - 6x^2)
as x approaches negative infinity

is the answer positive infinity, if not then can you explain
Divide through by $\displaystyle x^2$ to get $\displaystyle \frac{3x+3}{\frac{1}{x}-6}$.

Now as $\displaystyle x\to-\infty$ it should clear we get positive numbers.