- #1
skam74
- 4
- 0
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.
[tex]\underset{x\rightarrow0}{\lim}\frac{x-xcosx}{sin^{2}3x}[/tex]
[tex]\sin ^{2}x = \frac{1-\cos 2x}{2}[/tex]
[tex]\frac{x}{\frac{y}{z}}=x\cdot \frac{z}{y}[/tex]
[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]
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]