limits! : )

**iz1hp** · October 2nd 2008, 12:49 PM

Find the following limits:

$ \lim_{x \to 0} \frac{tan 2x}{3x cosx} $

$ \lim_{x \to 0} x^3 sin \frac{2}{x} $

mhmmmm help kthx : )

**icemanfan** · October 2nd 2008, 01:02 PM

1. $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$

$\frac{\tan 2x}{3x \cos x} =$

$\frac{2 \tan x}{3x \cos x (1 - \tan^2 x)} =$

$\frac{2 \sin x}{3x \cos^2 x (1 - \tan^2 x)} =$

$\frac{2}{3} \cdot \frac{\sin x}{x} \cdot \frac{1}{\cos^2 x (1 - \tan^2 x)}$

The left limit is 2/3, the middle limit is 1, and the right limit is 1. Therefore, the limit of the product is 2/3.

2. $x^3 \sin \left(\frac{2}{x}\right)$

The limit of x^3 is 0, and the sine function is bounded between -1 and 1, so the limit is 0.