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Math Help - Limit with sin

  1. #1
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    Limit with sin

    evaluate the limit of (sin(3+x)^2 - sin9) / x
    as x approaches zero
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  2. #2
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    \lim_{x \to 0} \frac{\sin^{2}(3+x) - \sin 9}{x}

    Is this your limit? Seems like an odd question but if so, the limit does not exist:

     = \lim_{x \to 0} \left(\frac{\sin^{2}(3+x)}{x} - \frac{\sin 9}{x}\right) \: \: = \: \: \lim_{x \to 0} \frac{\sin^{2}(3+x)}{x} - {\color{blue} \lim_{x\to 0} \frac{\sin 9}{x}}

    since what is highlighted in blue does not have a limit (it is comparable to \lim_{x \to 0} \frac{1}{x} where \lim_{x \to 0^{+}} \frac{1}{x} \neq \lim_{x \to 0^{-}} \frac{1}{x})
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  3. #3
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    Think derivative! Using the limit definition of derivative of \sin(x^2) at x=3.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Plato View Post
    Think derivative! Using the limit definition of derivative of \sin(x^2) at x=3.
    Yawn....why do it the cool way when you have L'hopital's?

    You are right..that is a better method but more difficult to explain
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    You all should see this for what it is.
    \left. {\frac{{d\sin (x^2 )}}{{dx}}} \right|_{x = 3}  = \lim _{x \to 0} \frac{{\sin \left[ {(3 + x)^2 } \right] - \sin \left[ {(3)^2 } \right]}}{x} = 6\cos (9)
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    Quote Originally Posted by Mathstud28 View Post
    Yawn....why do it the cool way when you have L'hopital's?
    You do know that the is a considerable number of mathematicians in the calculus reform movement who think that l’Hospital’s rule should be dropped “freshman” calculus. I remember the horror I felt upon hearing that proposed while attending a conference on ‘calculus instruction’ in 1988. In the years since I have come to embrace that point. This very thread proves the correctness of the point.
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  7. #7
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    Quote Originally Posted by Plato View Post
    You all should see this for what it is.
    \left. {\frac{{d\sin (x^2 )}}{{dx}}} \right|_{x = 3} = \lim _{x \to 0} \frac{{\sin \left[ {(3 + x)^2 } \right] - \sin \left[ {(3)^2 } \right]}}{x} = 6\cos (9)
    im still confused as to how that works out to 6cos9
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Plato View Post
    You do know that the is a considerable number of mathematicians in the calculus reform movement who think that l’Hospital’s rule should be dropped “freshman” calculus. I remember the horror I felt upon hearing that proposed while attending a conference on ‘calculus instruction’ in 1988. In the years since I have come to embrace that point. This very thread proves the correctness of the point.
    I agree that you should not use L'hopital's rule excessively but you should learn it first I think so that when you are unsure of how to apply a more creative method you have a backup plan
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  9. #9
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    hmm
    i dont know hopitals rule yet but this reminds me of something

    find the derivatives using first principles
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  10. #10
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    Quote Originally Posted by finch41 View Post
    hmm
    i dont know hopitals rule yet but this reminds me of something
    find the derivatives using first principles
    That is exactly what Plato did. He used the definition of the derivative.
    So use,
    \lim_{x \to 0} \frac{f(x+a) - f(a)}{x} = f'(a)

    Use f(x) = \sin{x^2} and a=3.

    Mathstud28 is obsessed with L'hopitals for now. When you study it , you will be too
    It has an irresistible charm of killing of limit problems instantaneously.

    Quote Originally Posted by Mathstud28
    I agree that you should not use L'hopital's rule excessively but you should learn it first I think so that when you are unsure of how to apply a more creative method you have a backup plan
    I agree
    But when you do so, also tell the alternative cool method to the poster. After all thats what they have to write in the exam
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  11. #11
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    Quote Originally Posted by Isomorphism View Post
    That is exactly what Plato did. He used the definition of the derivative.
    So use,
    \lim_{x \to 0} \frac{f(x+a) - f(a)}{x} = f'(a)

    Use f(x) = \sin{x^2} and a=3.

    Mathstud28 is obsessed with L'hopitals for now. When you study it , you will be too
    It has an irresistible charm of killing of limit problems instantaneously.


    I agree
    But when you do so, also tell the alternative cool method to the poster. After all thats what they have to write in the exam

    oh i dont particularly know how to do it using first principles...
    i haven't worked with these types of problems much at all yet
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by finch41 View Post
    oh i dont particularly know how to do it using first principles...
    i haven't worked with these types of problems much at all yet
    Isomorphism showed you

    Apply the knowledge gained from this example

    \lim_{x\to{0}}\frac{arctan(x)}{x}

    Direct substitution yields \frac{0}{0}

    YOu have two routes...L'hopital's or this method

    THe definition of a derivative evaluated at a point c is f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}

    Now this looks an awful lot like that...since arctan(0)=0

    This can be rewritten as \lim_{x\to{0}}\frac{arctan(x)-arctan(0)}{x-0}

    which is exactly like that formula I just showed you


    so we have the right side of the formul all we need is the left side

    So in this case it would be f'(0)=\lim_{x\to{0}}\frac{arctan(x)-arctan(0)}{x-0}

    and since f(x)=arctan(x)
    we know that f'(x)=\frac{1}{1+x^2}

    so f'(0)=\frac{1}{1+0^2}=1

    thus the limit is one
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Isomorphism View Post
    That is exactly what Plato did. He used the definition of the derivative.
    So use,
    \lim_{x \to 0} \frac{f(x+a) - f(a)}{x} = f'(a)

    Use f(x) = \sin{x^2} and a=3.

    Mathstud28 is obsessed with L'hopitals for now. When you study it , you will be too
    It has an irresistible charm of killing of limit problems instantaneously.


    I agree
    But when you do so, also tell the alternative cool method to the poster. After all thats what they have to write in the exam
    Happy now? ...I even came up with an example I have never seen before...hooray! You are right...sometiems I just get lazy and go "L'hopital's" automatically
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  14. #14
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    so how do you differentiate sin (x^2) ?
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  15. #15
    Moo
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    Hello,

    Quote Originally Posted by Mathstud28 View Post
    Yawn....why do it the cool way when you have L'hopital's?

    Because the text states to use the definition of the derivative ?
    Plus, it's really boring to always see that you do the apology of this rule... If someone is asked to do a way, then he has to do this way if it is that clear... doesn't he ?

    You are right..that is a better method but more difficult to explain
    So why do you bother if you don't explain ?

    Quote Originally Posted by finch41 View Post
    so how do you differentiate sin (x^2) ?
    Using the chain rule :

    \left(\sin(f(x))\right)'=f'(x)\cos(f(x))
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