Homework Help: 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.

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