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Math Help - show function is continuous

  1. #1
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    show function is continuous

    hi, the only instruction is to show the function is continuous. i know the epsilon delta definition, but when that is required, usually it is explicitly mentioned. so, i am thinking i need to use a different idea. the function is defined as: arcsin(x)/x when x=/=0, and 1 when x=0. so, i am thinking about what method to use. should i show that when arcsin(x)/x approaches 0, the limit is 1? is this a sufficient condition? should i show that arcsinx/x is continuous elsewhere? or can i just state this (and possibly why?)
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    Re: show function is continuous

    You have to show :

    \lim_{x \to 0^{-}}\frac{\arcsin(x)}{x}=\lim_{x \to 0^{+}}\frac{\arcsin(x)}{x}=1
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    Re: show function is continuous

    right. so. how can i do this?
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    Re: show function is continuous

    Quote Originally Posted by learning View Post
    right. so. how can i do this?
    L'Hopital's rule
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    Re: show function is continuous

    is l'opital essential here, i will be showing what with this? i understand the conditions are met for it, but, thn what? lopital shows differentiabiliy, which is the next question. so i need to show continuity here in a way which does not show differentiability.
    Last edited by learning; March 14th 2013 at 02:32 AM.
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    Re: show function is continuous

    i now have to show that it is in fact differentiable at 0. i am thinking again l opital here. but showing the conditions apply?
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    Re: show function is continuous

    To be differential at a point, the function needs to be continuous at that point AND reasonably smooth. To be smooth, that means that the gradients don't make sudden jumps. So the derivative ALSO needs to be continuous at that point.
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    Re: show function is continuous

    i need to show that (arcsinx/x - 1 )/ x goes to 0 as x goes to 0. it should be l ' hopitalable but arcsin0/0 - 1 has to equal 0 for that to be so. in the original function, the function is defined as 1 at x=0, can i use this here to show the condition for l' hopital? even though its technically a new function? can i plug in that earlier definition?
    Last edited by learning; March 14th 2013 at 04:09 AM.
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    Re: show function is continuous

    L'Hopital's rule is used for the limit of a fraction - when you get \frac{0}{0} or \frac{\infty}{\infty}, you can take the derivative of the numerator and denominator and try again. So you have arcsin(x) for the numerator and x for the denominator. If you plug in the derivatives and set x=0, you should get the answer.

    - Hollywood
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    Re: show function is continuous

    in terms of showing differentiability at 0. would l'hopital be appropriate again.
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    Re: show function is continuous

    To show it's differentiable, you have a different limit to prove: \lim_{h\to{0}}\frac{\frac{\arcsin{h}}{h}-1}{h}.

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    Re: show function is continuous

    absolutely, i know, but then if you use l'opital here. my issue is, can you use the fact that arcsinx/x was defined as 1 at 0, to say that the numerator is 0 at h=0. (thus l' hopital is valid).? or since its a 'new' function you can't do this. l'hopital
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    Re: show function is continuous

    i know what i have to do just not how to do it
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    Re: show function is continuous

    You need to use L'Hopital's rule twice:

    \lim_{h\to{0}}\frac{\frac{\arcsin{h}}{h}-1}{h}=

    \lim_{h\to{0}}\frac{\arcsin{h}-h}{h^2}=

    \lim_{h\to{0}}\frac{\frac{1}{\sqrt{1-h^2}}-1}{2h}=

    \lim_{h\to{0}}\frac{\frac{h}{(1-h^2)^\frac{3}{2}}}{2}=0

    Since the limit exists, the function is differentiable.

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    Re: show function is continuous

    right, thats what i thought. but its the first line i wasnt sure about. whether i need to multiply through by h first, or if i could use the definition of the function to 'replace' the arcsinh/h with a 1 (since the function is defined as 1 at 0, therefore the function - 1 would be defined as 0 at 0). though., the net result is the same. thats actually what i did. to be honest. what you wrote above is exactly what i did. thanks for the confirmation.
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