Limit $\lim _{ \theta \to 0 }{ \frac { cos2\theta -cos\theta }{ \theta } } $ $$\lim _{ \theta \to 0 }{ \frac { cos2\theta -cos\theta  }{ \theta  }  } $$
Steps I took:
$$\lim _{ \theta \rightarrow 0 }{ \frac { 1-2sin^{ 2 }\theta -cos\theta  }{ \theta  }  } =$$
$$\lim _{ \theta \rightarrow 0 }{ \frac { -2sin^{ 2 }\theta  }{ \theta  }  } +\lim _{ \theta \rightarrow 0 }{ \frac { 1-cos\theta  }{ \theta  }  } $$
$$\lim _{ \theta \rightarrow 0 }{ \frac { 1-cos\theta  }{ \theta  } =0 } $$
$$\lim _{ \theta \rightarrow 0 }{ \frac { -2(sin\theta )(sin\theta ) }{ \theta  } = } $$
$$\lim _{ \theta \rightarrow 0 }{ { \quad -2(sin\theta ) }\cdot 1=0 } $$
$$0+0=0$$
Something seems off about the way I went about this but I can't figure it out.
 A: L'Hospital is very efficient here: just by inspection, $\dfrac{-0+0}1$.
Also, as the numerator is an even smooth function, it must be quadratic in $\theta$, hence $0$.
A: What you'll do if you need to calcualte $\dfrac{\cos\pi\theta-\cos\dfrac32\theta}{\theta}$?
Following method encompasses such possibilities as well.
Use Prosthaphaeresis Formula,
$$\cos(2m\theta)-\cos(2n\theta)=-2\sin(m+n)\theta\sin(m-n)\theta$$
and $\lim_{h\to0}\dfrac{\sin h}h=1$
Here $2m=2,2n=1$
A: $$\lim _{ \theta \rightarrow 0 }{ \frac { 1-cos\theta  }{ \theta }}=\lim _{ \theta \rightarrow 0 }{ \frac { 2 \sin^2 \frac{\theta}{2} }{ \theta  } =0 }$$
A: your proof is correct. but if you are going to use $\lim_{\theta \to 0}\frac{1-\cos \theta}{\theta} = 0,$ you could have split $\cos(2\theta) - \cos \theta$ as $(1-\cos \theta) -(1 - \cos 2 \theta)$ at the beginning itself.
A: Just to give you an other way to calculate your limit among these good answers
$\cos\theta=1-\frac{\theta^2}{2}+ o(\theta^2)$ if $\theta\to 0$, and thus
$$\lim_{\theta\to 0}\frac{\cos 2\theta-\cos \theta}{\theta}=\lim_{\theta\to 0}\frac{1-\frac{4\theta^2}{2}-1+\frac{\theta^2}{2}}{\theta}=\lim_{\theta\to 0}\frac{-3\theta}{2}=0.$$
A: You can keep the symmetry:
$$\frac{\cos2\theta-\cos\theta}\theta=\frac{\cos2\theta-1}{\theta}-\frac{\cos\theta-1}\theta=\frac{2\sin^2\theta}{\theta}-\frac{2\sin^2\frac\theta2}{\theta}\to0-0$$
