# Basic Limit Involving x-xcosx and sin^2(3x)

1. Sep 1, 2011

### skam74

My answer is correct, but I feel like I might've taken an illegitimate route to get there. Can anyone verify? Many thanks in advance.

1. The problem statement, all variables and given/known data
$$\underset{x\rightarrow0}{\lim}\frac{x-xcosx}{sin^{2}3x}$$

2. Relevant equations
$$\sin ^{2}x = \frac{1-\cos 2x}{2}$$
$$\frac{x}{\frac{y}{z}}=x\cdot \frac{z}{y}$$

3. The attempt at a solution
$$\underset{x\rightarrow0}{\lim}\frac{x-xcosx}{sin^{2}3x}$$
$$=\underset{x\rightarrow0}{\lim}\frac{(x-x)cosx}{(\cfrac{1-cos(6x)}{2})}$$
$$=\underset{x\rightarrow0}{\lim}\frac{(x-x)cosx}{1}\cdot\frac{2}{1-cos(6x)}$$
$$=\underset{x\rightarrow0}{\lim}2((x-x)cosx)\cdot(1-cos(6x))$$
$$=2((0)cos0)\cdot(1-cos0)$$
$$=2(0\cdot1)\cdot(1-1)$$
$$=0$$

2. Sep 1, 2011

### Ray Vickson

The second line of (3) is wrong: you cannot write x - x*cos(x) as (x-x)*cos(x). However, you can write it as x*2*sin(x/2)^2, so your function is 2*x*[sin(x/2)/sin(3x)]^2.

RGV

3. Sep 1, 2011

### skam74

Never mind I see it now:

sin^2(3x) = sin^2(2x + x) = [cos(2x)sin(x) + sin(2x)cos(x)]^2

which ends up giving 0/4 = 0

4. Sep 2, 2011

### skam74

Just for fun I solved it using your idea as well:

$$\lim_{x\rightarrow 0}\frac{x-xcosx}{sin^{2}3x}$$
$$=\lim_{x\rightarrow 0}\frac{x(1-cosx)}{sin^23x}$$
$$=\lim_{x\rightarrow 0}\frac{x(2sin^2(\frac{x}{2}))}{sin^23x}$$
$$=\lim_{x\rightarrow 0}\frac{xsinx}{\frac{2}{sin^23x}}$$
$$=\lim_{x\rightarrow 0}\frac{xsinx}{2}\cdot \frac{1}{sin^23x}$$
$$=\lim_{x\rightarrow 0}\frac{xsinx\cdot csc3x}{2}$$
$$=\frac{0}{2}$$
$$=0$$

5. Sep 2, 2011

### vela

Staff Emeritus
Your last attempt has numerous errors in it, and I don't see how what you wrote in post #3 helps in evaluating the limit.

6. Sep 2, 2011

### skam74

Oops. Can you point me towards my errors in my last attempt?

7. Sep 2, 2011

### vela

Staff Emeritus
Going from the third to fourth lines, fifth to sixth lines. Finally, when evaluating the limit at the end, you seemed to pull a 0 out of nowhere. Cosecant diverges as x goes to 0, so you have an indeterminate form.