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

[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.

Homework Statement

\underset{x\rightarrow0}{\lim}\frac{x-xcosx}{sin^{2}3x}

Homework Equations

\sin ^{2}x = \frac{1-\cos 2x}{2}
\frac{x}{\frac{y}{z}}=x\cdot \frac{z}{y}

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

 

[Ray Vickson]

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.

Homework Statement

\underset{x\rightarrow0}{\lim}\frac{x-xcosx}{sin^{2}3x}

Homework Equations

\sin ^{2}x = \frac{1-\cos 2x}{2}
\frac{x}{\frac{y}{z}}=x\cdot \frac{z}{y}

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

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

 

[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

 

[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

 

[vela]

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.

 

[skam74]

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.

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

 

[vela]

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.