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

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  • #1
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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


[tex]\underset{x\rightarrow0}{\lim}\frac{x-xcosx}{sin^{2}3x}[/tex]

Homework Equations


[tex]\sin ^{2}x = \frac{1-\cos 2x}{2}[/tex]
[tex]\frac{x}{\frac{y}{z}}=x\cdot \frac{z}{y}[/tex]



The Attempt at a Solution


[tex]\underset{x\rightarrow0}{\lim}\frac{x-xcosx}{sin^{2}3x}[/tex]
[tex]=\underset{x\rightarrow0}{\lim}\frac{(x-x)cosx}{(\cfrac{1-cos(6x)}{2})}[/tex]
[tex]=\underset{x\rightarrow0}{\lim}\frac{(x-x)cosx}{1}\cdot\frac{2}{1-cos(6x)}[/tex]
[tex]=\underset{x\rightarrow0}{\lim}2((x-x)cosx)\cdot(1-cos(6x))[/tex]
[tex]=2((0)cos0)\cdot(1-cos0)[/tex]
[tex]=2(0\cdot1)\cdot(1-1)[/tex]
[tex]=0[/tex]
 

Answers and Replies

  • #2
Ray Vickson
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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


[tex]\underset{x\rightarrow0}{\lim}\frac{x-xcosx}{sin^{2}3x}[/tex]

Homework Equations


[tex]\sin ^{2}x = \frac{1-\cos 2x}{2}[/tex]
[tex]\frac{x}{\frac{y}{z}}=x\cdot \frac{z}{y}[/tex]



The Attempt at a Solution


[tex]\underset{x\rightarrow0}{\lim}\frac{x-xcosx}{sin^{2}3x}[/tex]
[tex]=\underset{x\rightarrow0}{\lim}\frac{(x-x)cosx}{(\cfrac{1-cos(6x)}{2})}[/tex]
[tex]=\underset{x\rightarrow0}{\lim}\frac{(x-x)cosx}{1}\cdot\frac{2}{1-cos(6x)}[/tex]
[tex]=\underset{x\rightarrow0}{\lim}2((x-x)cosx)\cdot(1-cos(6x))[/tex]
[tex]=2((0)cos0)\cdot(1-cos0)[/tex]
[tex]=2(0\cdot1)\cdot(1-1)[/tex]
[tex]=0[/tex]

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

[tex]\lim_{x\rightarrow 0}\frac{x-xcosx}{sin^{2}3x}[/tex]
[tex]=\lim_{x\rightarrow 0}\frac{x(1-cosx)}{sin^23x}[/tex]
[tex]=\lim_{x\rightarrow 0}\frac{x(2sin^2(\frac{x}{2}))}{sin^23x}[/tex]
[tex]=\lim_{x\rightarrow 0}\frac{xsinx}{\frac{2}{sin^23x}}[/tex]
[tex]=\lim_{x\rightarrow 0}\frac{xsinx}{2}\cdot \frac{1}{sin^23x}[/tex]
[tex]=\lim_{x\rightarrow 0}\frac{xsinx\cdot csc3x}{2}[/tex]
[tex]=\frac{0}{2}[/tex]
[tex]=0[/tex]
 
  • #5
vela
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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
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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?
 
  • #7
vela
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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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