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Math Help - Classic limit

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Classic limit

    I came up with seven real ways of doing this limit...and one stupid way ...how many can you come up with? It's simple I know...but let see if we can get eight?

    \lim_{\Delta{x}\to{0}}\frac{1-\cos(\Delta{x})}{\Delta{x}}
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  2. #2
    Super Member PaulRS's Avatar
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    1.

    1-\cos(\Delta x)=2\cdot{\sin^2\left(\frac{\Delta x}{2}\right)}

    and \frac{\sin(\Delta x)}{\Delta x}\rightarrow{1} as \Delta x\rightarrow{0} (*)

    2.

    Multiply numerator and denominator by: 1+\cos(\Delta x)

    we have: 1-\cos^2(\Delta x)=\sin^2(\Delta x) and then apply (*)



    3. Definition of f'(0) where f(x)=1-\cos(x)

    ( this is not valid in my opinion)

    4. \int_0^1\sin(\Delta x\cdot{y})dy=\frac{1-\cos(\Delta x)}{\Delta x}

    So if it's uniformly convergent (you'd have to go over that more carefully)
    we have: \lim_{\Delta x\rightarrow{0}}\int_0^1\sin(\Delta x\cdot{y})dy=\int_0^1\lim_{\Delta x\rightarrow{0}}\sin(\Delta x\cdot{y})dy=0

    5. assuming it exists

    <br />
L = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{1 - \cos \left( {\Delta x} \right)}}<br />
{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos \left( {\Delta x} \right) - \cos ^2 \left( {\Delta x} \right)}}<br />
{{\Delta x}}<br />
since <br />
\cos \left( {\Delta x} \right) \to 1<br />

    Then <br />
L = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos \left( {\Delta x} \right) - 1 + \sin ^2 \left( {\Delta x} \right)}}<br />
{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos \left( {\Delta x} \right) - 1}}<br />
{{\Delta x}} =  - L<br />
thus: L=0

    You counted applying the Power Series expansion of cos(x), right?
    Last edited by PaulRS; May 11th 2008 at 07:02 PM.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    1.

    1-\cos(\Delta x)=2\cdot{\sin^2\left(\frac{\Delta x}{2}\right)}

    and \frac{\sin(\Delta x)}{\Delta x}\rightarrow{1} as \Delta x\rightarrow{0} (*)

    2.

    Multiply numerator and denominator by: 1+\cos(\Delta x)

    we have: 1-\cos^2(\Delta x)=\sin^2(\Delta x) and then apply (*)



    3. Definition of f'(0) where f(x)=1-\cos(x)

    ( this is not valid in my opinion)

    4. \int_0^1\sin(\Delta x\cdot{y})dy=\frac{1-\cos(\Delta x)}{\Delta x}

    So if it's uniformly convergent (you'd have to go over that more carefully)
    we have: \lim_{\Delta x\rightarrow{0}}\int_0^1\sin(\Delta x\cdot{y})dy=\int_0^1\lim_{\Delta x\rightarrow{0}}\sin(\Delta x\cdot{y})dy=0
    Missing three
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  4. #4
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    There is nothing amazing about this problem.
    Because mathematicaly cosine is not a relationship between the sides of a right-triangle rather it is simply \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}.

    With that fact the problem is trivial.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    There is nothing amazing about this problem.
    Because mathematicaly cosine is not a relationship between the sides of a right-triangle rather it is simply \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}.

    With that fact the problem is trivial.
    O come on don't be a spoil sport ...I know its trivial its a simple limit but I want to see if anyone finds one I haven't (I am sure someone will)....if its so trivial why dont you present 3-4 other methods?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Another one, assuming it exists

    <br />
L = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{1 - \cos \left( {\Delta x} \right)}}<br />
{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos \left( {\Delta x} \right) - \cos ^2 \left( {\Delta x} \right)}}<br />
{{\Delta x}}<br />
since <br />
\cos \left( {\Delta x} \right) \to 1<br />

    Then <br />
L = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos \left( {\Delta x} \right) - 1 + \sin ^2 \left( {\Delta x} \right)}}<br />
{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos \left( {\Delta x} \right) - 1}}<br />
{{\Delta x}} = - L<br />
thus: L=0

    You counted applying the Power Series expansion of cos(x), right?
    Believe it or not that is one that I actually did that method ...and as for your question...ahem...I don't know if you know this...but I am actually kind of famous on this website...I use power series at completely inappropriate places...of course I used and counted it here
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  7. #7
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Believe it or not that is one that I actually did that method ...and as for your question...ahem...I don't know if you know this...but I am actually kind of famous on this website...I use power series at completely inappropriate places...of course I used and counted it here
    On the same fashion:

    <br />
L = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( x \right)}}<br />
{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos ^2 \left( x \right) - \cos ^3 \left( x \right)}}<br />
{x}<br />

    <br /> <br />
L = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos ^3 \left( x \right)}}<br />
{x} - \mathop {\lim }\limits_{x \to 0} \frac{{\sin ^2 \left( x \right)}}<br />
{x} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos ^3 \left( x \right)}}<br />
{x}<br /> <br />
(*)

    <br />
L = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( x \right)}}<br />
{x} \cdot \left[ {1 + \cos \left( x \right) + \cos ^2 \left( x \right)} \right]<br />
=3\cdot{L}<br /> <br />
thus L=0

    or (*) <br />
\cos ^3 \left( x \right) = \frac{{\cos \left( {3x} \right) + 3 \cdot \cos \left( x \right)}}<br />
{4}<br />

    <br />
L = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos ^3 \left( x \right)}}<br />
{x} = \tfrac{1}<br />
{4} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {3x} \right) + 3 \cdot \left[ {1 - \cos \left( x \right)} \right]}}<br />
{x}<br />

    <br />
L = \tfrac{3}<br />
{4} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {3x} \right)}}<br />
{{3x}} + \tfrac{3}<br />
{4} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( x \right)}}<br />
{x} = \frac{3}<br />
{2} \cdot L<br /> <br />
thus L=0

    Ok, enough, I'm going to now
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    On the same fashion:

    <br />
L = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( x \right)}}<br />
{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos ^2 \left( x \right) - \cos ^3 \left( x \right)}}<br />
{x}<br />

    <br /> <br />
L = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos ^3 \left( x \right)}}<br />
{x} - \mathop {\lim }\limits_{x \to 0} \frac{{\sin ^2 \left( x \right)}}<br />
{x} = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos ^3 \left( x \right)}}<br />
{x}<br /> <br />
(*)

    <br />
L = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( x \right)}}<br />
{x} \cdot \left[ {1 + \cos \left( x \right) + \cos ^2 \left( x \right)} \right]<br />
=3\cdot{L}<br /> <br />
thus L=0

    or (*) <br />
\cos ^3 \left( x \right) = \frac{{\cos \left( {3x} \right) + 3 \cdot \cos \left( x \right)}}<br />
{4}<br />

    <br />
L = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos ^3 \left( x \right)}}<br />
{x} = \tfrac{1}<br />
{4} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {3x} \right) + 3 \cdot \left[ {1 - \cos \left( x \right)} \right]}}<br />
{x}<br />

    <br />
L = \tfrac{3}<br />
{4} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {3x} \right)}}<br />
{{3x}} + \tfrac{3}<br />
{4} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( x \right)}}<br />
{x} = \frac{3}<br />
{2} \cdot L<br /> <br />
thus L=0

    Ok, enough, I'm going to now
    That is a new one haha....long-winded but cool nonetheless
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  9. #9
    o_O
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    Quote Originally Posted by ThePerfectHacker View Post
    There is nothing amazing about this problem.
    Because mathematicaly cosine is not a relationship between the sides of a right-triangle rather it is simply \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}.
    What do you mean? Can you elaborate a little. Trigonometric functions are functions of an angle aren't they?

    ----

    Wiki'd it and it seems that they're defined by their taylor series o.O.
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  10. #10
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    Quote Originally Posted by o_O View Post
    What do you mean? Can you elaborate a little. Trigonometric functions are functions of an angle aren't they?
    What you need to understand is that mathematics falls into two categories. The formal and the intuitive. When mathematics is developed as a theory, for example, analysis, it is treated on a very formal manner. However, mathematics also has some nice intuitive interpretations behind it, such as, geometry (one of the reasons why I like complex analysis, it is very formal, and yet it can be very geometric). The sine function is an example. It is an important function, and when defined formally we simply define it as an infinite series. But it happens to be that the sine function has geometric interpretations behind it as well.

    Think of it this way. Suppose you were teaching a course in analysis. And you wanted to define the sine function. How would you define it? If you say "it is the ratio of the opposite to the the hypotenuse", that is no good. Because it is not formal enough. You need to define what it means "opposite" and "hypotenuse" but before doing that you need to define what it means "triangle" and before doing that you need to define what it means "length". All these words are not strict mathematical words. They are more geometric expressions. In high school you learn the sine function in terms of geometry but on a more formal level the sine function is just a power series.
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