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(cos θ - 1) / θ = 1


{$$ \frac {\cos \theta - 1} {\theta} \tag{Fig. 1} $$}

Prove

{$$ \lim_{\theta\rightarrow0} {\frac{\cos\theta - 1}{\theta}} = 0 $$}

General Approach

This proof use trigonometry identities, a previous result

{$$ \lim_{\theta\rightarrow0} {\frac{\sin\theta}{\theta}} = 1 $$}

which is demonstrated here, which will be used to rid the denominator of θ and the product of limits.

{$$ \begin{gather} \lim_{\theta\rightarrow0} {\frac{\cos\theta - 1}{\theta}} &= \cr \lim_{\theta\rightarrow0} {\frac{(\cos\theta - 1)(\cos\theta + 1)}{\theta(\cos\theta + 1)}} &= \cr \lim_{\theta\rightarrow0} {\frac{\cos^2\theta - 1}{\theta(\cos\theta + 1)}} & \end{gather} $$}

Argument: From the trig identity sin²θ + cos²θ = 1, we see cos²θ - 1 = -sin²θ.

{$$ \begin{gather}\lim_{\theta\rightarrow0} {\frac{\cos^2\theta - 1}{\theta(\cos\theta + 1)}} &= \cr \lim_{\theta\rightarrow0} {\frac{-\sin^2\theta}{\theta(\cos\theta + 1)}} &= \cr \lim_{\theta\rightarrow0} \left( \large{-1} \right) \left( {\frac{\sin\theta}{\theta}} \right) \left( \sin\theta \right) \left( {\frac{1}{\cos\theta + 1}} \right) &= \cr \left(\lim_{\theta\rightarrow0} \large{-1} \right) \left(\lim_{\theta\rightarrow0} {\frac{\sin\theta}{\theta}} \right) \left( \lim_{\theta\rightarrow0} {\sin\theta} \right) \left( \lim_{\theta\rightarrow0} {\frac{1}{\cos\theta + 1}} \right) \end{gather} $$}

Argument: The limit of products is the product of limits.

{$$ \begin{gather} \left( \lim_{\theta\rightarrow0} \large{-1} \right) \left( \lim_{\theta\rightarrow0} {\frac{\sin\theta}{\theta}} \right) \left( \lim_{\theta\rightarrow0} {\sin\theta} \right) \left( \lim_{\theta\rightarrow0} {\frac{1}{\cos\theta + 1}} \right) = \cr \left( \large{-1} \right) \left( \large{1} \right) \left( \large{0} \right) \left( \lim_{\theta\rightarrow0} {\frac{1}{\cos\theta + 1}} \right) = \large{0} \cr \therefore \boxed{\lim_{\theta\rightarrow0} {\frac{\cos\theta - 1}{\theta}} = 0} \end{gather} $$}

Argument: The limit of the constant -1 is the constant. {$$ \lim_{\theta\rightarrow0} {\frac{\sin\theta}{\theta}} = 1 $$} is a previous result, demonstrated here, since there is no denominator {$$ \lim_{\theta\rightarrow0} \sin\theta $$} can be found by substitution to be 0,

{$$ \lim_{\theta\rightarrow0} {\frac{1}{\cos\theta + 1}} $$} is defined because cosθ is near 1 (not near -1), so the denominator is not 0. In fact this limit is ½ but its value does not matter. Since one factor is 0, the whole product is 0.


Graph:

  1. FooPlot: Online graphing calculator and function plotter

Sources:

Wikisource

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Category: Math


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This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

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August 06, 2017

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