Originally Posted by

**kbartlett** For the 1st one, $\displaystyle 4x^2-3x^3=x^3(\frac{4}{x}-3)$, and as $\displaystyle x\to\infty$ it is obvious $\displaystyle x^3\to\infty$, and $\displaystyle \frac{4}{x}-3\to\--3$ so the limit is $\displaystyle -\infty$ is my method right?

Using l'hopitals rule for the second i got the limit as 1/2.

For the 4th one, i think the limit is 0 because $\displaystyle \frac{1}{x}\to\infty$ as $\displaystyle x\to\infty$ is that right?

could any1 confirm if my answers are correct?