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$$\lim _{ \theta \to 0 }{ \frac { cos2\theta -cos\theta }{ \theta } } $$

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.

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  • $\begingroup$ Fixed. Is the rest correct? $\endgroup$ – Cherry_Developer Jan 20 '15 at 7:05
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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.

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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 $0$.

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$$\lim _{ \theta \rightarrow 0 }{ \frac { 1-cos\theta }{ \theta }}=\lim _{ \theta \rightarrow 0 }{ \frac { 2 \sin^2 \frac{\theta}{2} }{ \theta } =0 }$$

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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 $2m=2,2n=1$

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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 $$\lim_{\theta\to 0}\frac{\cos 2\theta-\cos \theta}{\theta}=\lim_{\theta\to 0}\frac{1-\frac{4\theta^2}{2}-1+\frac{\theta^2}{2}}{\theta}=\lim_{\theta\to 0}\frac{-3\theta}{2}=0.$$

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You can keep the symmetry:

$$\frac{\cos2\theta-\cos\theta}\theta=\frac{\cos2\theta-1}{\theta}-\frac{\cos\theta-1}\theta=\frac{2\sin^2\theta}{\theta}-\frac{2\sin^2\frac\theta2}{\theta}\to0-0$$

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