1. ## limits! : )

Find the following limits:

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

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

mhmmmm help kthx : )

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

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

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

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

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

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

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

3. ok so like

$\displaystyle \lim_{x \to 0^-} \frac{5}{csc x}$

that equals 0?
because..

$\displaystyle \frac{5}{csc x} = 5 \cdot sinx = 5 sin(0) = 5 \cdot 0 = 0$

4. Originally Posted by iz1hp
ok so like

$\displaystyle \lim_{x \to 0^-} \frac{5}{csc x}$

that equals 0?
because..

$\displaystyle \frac{5}{csc x} = 5 \cdot sinx = 5 sin(0) = 5 \cdot 0 = 0$
Correct.

5. excellent : ) im grasping the concept hahaha thanks!