For the first one:
\(\displaystyle \L \frac{x^3}{tan^3(x)} = \frac{x^3}{\frac{sin^3(x)}{cos^3(x)}} = (\frac{x}{sin(x)})(\frac{x}{sin(x)})(\frac{x}{sin(x)})cos^3(x)\)
By AOL (Algebra Of Limits), you can take the limit of each factor seperately and multiply them.
For the second:
\(\displaystyle \L
(\frac{x}{tan(2x)})^3 = \frac{x^3}{\frac{sin^3(2x)}{cos^3(2x)}} = (\frac{x}{sin2x})(\frac{x}{sin2x})(\frac{x}{sin2x})cos^3(2x)\)
Now multiply each \(\displaystyle (\frac{x}{sin(2x)})\) by 2. You are actually multiplying the entire term by 8 (2*2*2), so also divide by 8. This way you aren't actually changing the term, just making it look like something familiar:
\(\displaystyle \L
(\frac{x}{sin2x})(\frac{x}{sin2x})(\frac{x}{sin2x})cos^3(2x) = \frac{1}{8}(\frac{2x}{sin2x})(\frac{2x}{sin2x})(\frac{2x}{sin2x})cos^3(2x)\)
Hope that helps,
-Daon
