Originally Posted by

**bcahmel** I wanted to make sure I'm doing this right.

1. $\displaystyle \displaystyle\lim_{x \to\ 0} \frac{x^3}{sinx-x}$

Get 0/0, so can use L'hopital's rule.

$\displaystyle \displaystyle\frac{3x^2}{cosx-1}\rightarrow\frac{-6x}{sinx}\rightarrow\frac{-6}{cosx}$

Plugging a 0 into the last one gives -6. Is this technique right? **correct**

2. $\displaystyle \displaystyle\lim_{x \to \infty}\frac{x^2+4x}{9x^3+4}$.

I'm confused on how to approach an infinity one. How do I see both the numerator and denominator go to infinity(well, I guess I can tell if I graph). Can someone walk me through the steps? I would really appreciate it.