Question

lim (3-3cosx)/( 3x)
x->0

Answer 1

\[\lim_{x \rightarrow 0}\frac{ 3-3\cos x }{ 3x }\]
\[=\lim_{x \rightarrow 0}\frac{ 3\left( 1-\cos x \right) }{ 3x }=\lim_{x \rightarrow0}\frac{ 2 \sin ^2\frac{ x }{ 2 } }{ x*\frac{ x }{ 4 } }\times \frac{ x }{ 4 }\]
\[=2\lim_{x \rightarrow 0}\left( \frac{ \sin \frac{ x }{ 2 } }{ \frac{ x }{ 2 } } \right)^2*\lim_{x \rightarrow0} x\]
\[=\frac{ 1 }{ 2 }*1^2*0=0\]

Answer 2

correction second line from last write 2/4 in place of 2.

Answer 3

How'd it go from 3(1-cosx)/3x to 2sin^2(pi/2) in the first line

Answer 4

\[or \cos 2 y=\cos \left( y+y \right)=\cos y \cos y-\sin y \sin y\]
\[\cos 2y=\cos ^2y-\sin ^2y=1-\sin ^2y-\sin ^2y=1-2\sin ^2y\]
\[2\sin ^2y=1-\cos 2y\]
put 2y=x
y=x/2
\[2 \sin ^2 \frac{ x }{ 2 }=1-\cos x\]