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Epsilon Delta - limits watch

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    I'm really not understanding a thing here so would appreciate it if someone could explain it I've read both Wikipedia and my notes and they say pretty much the same thing, but I'm finding it quite hard to figure out what it actually means. I have:

    If \displaystyle K = \lim_{x\to x_0}[f(x)+g(x)]

    Then for any \epsilon > 0, \exists\delta>0 such that |K-f(x)-g(x)|<\epsilon for all 0<|x-x_0|<\delta

    Which, embarrassingly, is pretty much just a load of symbols to me
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    (Original post by Zygroth)
    I'm really not understanding a thing here so would appreciate it if someone could explain it I've read both Wikipedia and my notes and they say pretty much the same thing, but I'm finding it quite hard to figure out what it actually means. I have:

    If \displaystyle K = \lim_{x\to x_0}[f(x)+g(x)]

    Then for any \epsilon > 0, \exists\delta>0 such that |K-f(x)-g(x)|<\epsilon for all 0<|x-x_0|<\delta

    Which, embarrassingly, is pretty much just a load of symbols to me
    We say that \displaystyle K = \lim_{x\to x_0}[f(x)+g(x)] if for any \epsilon > 0(ie no matter how small epsilon is, we can find delta to satisfy it), then \exists\delta>0 (there is a delta, which is going to be the maximum difference between x and x_0) such that |K-f(x)-g(x)|<\epsilon for all 0<|x-x_0|<\delta(if we pick an x close enough to x_0, then f(x)+g(x) will be less than epsilon away from K is what this is saying in essence. The delta and epsilon are just quantifying quite 'how close' you have to be.)

    Make any more sense, or am I just being unhelpful?:p:
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    draw a picture.

    analysis is usually the proving of "obvious" statements which can be illustrated by pictures.
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    (Original post by Slumpy)
    Make any more sense, or am I just being unhelpful?:p:
    Lol, what is epsilon/delta though? Is there any use for this? It sounds a bit like just choosing random numbers closer and closer to where you want to find the limit...
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    (Original post by Zygroth)
    Lol, what is epsilon/delta though? Is there any use for this? It sounds a bit like just choosing random numbers closer and closer to where you want to find the limit...
    They're anything you want.
    As suggested above, draw a picture of a function. The limit existing just means that (roughly speaking), if we get closer and closer to a point on the x-axis, x_0, the y values will be getting closer and closer to some number(note, not necesarily the value of y at x_0, this is the definition of continuity). The epsilon is how close we want the y values to be to this limit, and the delta is how small a region around x_0 ensures this happens. So delta is normally a function of epsilon, in order to show it can be done for any epsilon.
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    So you probably know
    |x_n - L| < E for all n>N

    This says a term x_n gets closer to a limit L as n gets bigger.
    But what is "close" and "big"?
    There is no concrete answer for this, so if someone tells us what 'close' is, we can tell them what 'big' is. 'Closeness' is measured by epilson.
    This allows us to find an n that satsifies the n>N part.
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    (Original post by Slumpy)
    They're anything you want.
    As suggested above, draw a picture of a function. The limit existing just means that (roughly speaking), if we get closer and closer to a point on the x-axis, x_0, the y values will be getting closer and closer to some number(note, not necesarily the value of y at x_0, this is the definition of continuity). The epsilon is how close we want the y values to be to this limit, and the delta is how small a region around x_0 ensures this happens. So delta is normally a function of epsilon, in order to show it can be done for any epsilon.
    So if the limit exists, then the difference between the function and the limit (epsilon?) at some point close to it (delta?) will be not bigger than any value we choose? Would this be useful in finding the limit of something?

    (Original post by assmaster)
    So you probably know
    |x_n - L| < E for all n>N

    This says a term x_n gets closer to a limit L as n gets bigger.
    But what is "close" and "big"?
    There is no concrete answer for this, so if someone tells us what 'close' is, we can tell them what 'big' is. 'Closeness' is measured by epilson.
    This allows us to find an n that satsifies the n>N part.
    Not sure what E, n and N are... Can't say I've ever seen that o.O
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    (Original post by Zygroth)
    Not sure what E, n and N are... Can't say I've ever seen that o.O
    E was meant to be epsilon, sorry.
    n and N are just numbers - i.e. how 'big'
    Slumpy is explaining this better than me though, I feel.
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    (Original post by Zygroth)
    So if the limit exists, then the difference between the function and the limit (epsilon?) at some point close to it (delta?) will be not bigger than any value we choose? Would this be useful in finding the limit of something?



    Not sure what E, n and N are... Can't say I've ever seen that o.O
    The second bit is the definition of a convergence for a sequence.
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    (Original post by Zygroth)
    So if the limit exists, then the difference between the function and the limit (epsilon?) at some point close to it (delta?) will be not bigger than any value we choose? Would this be useful in finding the limit of something?
    You're mostly there;
    if the limit exsists, then we can make the difference between the function and the limit as small as we like. Epsilon is just a general number, so we say if we want the difference between the function and the limit to be less than it, how close to x_0 do we need to be? And how close we need to be to guarantee it is delta.
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    Stand explanation: It's a game! Somebody gives you some arbitrary small number called epsilon. Your job is, given this epsilon, to always be able to find another number (delta), which is how close x has to be to x_0 for you to be absolutely certain that f(x)+g(x) is epsilon close to K.
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    (Original post by majikthise)
    Stand explanation: It's a game! Somebody gives you some arbitrary small number called epsilon. Your job is, given this epsilon, to always be able to find another number (delta), which is how close x has to be to x_0 for you to be absolutely certain that f(x) (or in this case, f(x)+g(x)) is epsilon close to f(x_0).
    Careful-I don't think you mean f(x_0).
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    (Original post by Zygroth)
    I'm really not understanding a thing here so would appreciate it if someone could explain it I've read both Wikipedia and my notes and they say pretty much the same thing, but I'm finding it quite hard to figure out what it actually means. I have:

    If

    Then for any , such that for all

    Which, embarrassingly, is pretty much just a load of symbols to me
    Essentially this is the statement that if such a limit is satisfied, if we are given a 'distance' from the actual limit, denoted epsilon, we can pick a delta such that we are a distance of less than epsilon from the limit.

    Basically it states that we can get as arbitrarily close to the actual limit as we like (though not zero distance from it).
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    (Original post by studyboy)
    The second bit is the definition of a convergence for a sequence.
    Which bit? o.O I'll probably be asking questions about that when I get to it :p:

    (Original post by Slumpy)
    You're mostly there;
    if the limit exsists, then we can make the difference between the function and the limit as small as we like. Epsilon is just a general number, so we say if we want the difference between the function and the limit to be less than it, how close to x_0 do we need to be? And how close we need to be to guarantee it is delta.
    Aah, so we're choosing how close to the function we want to be and we find how far away x can be? So say the function is \dfrac{\sin 2x}{x}, and the limit is 2 as x -> 0. Epsilon is something we choose (e.g. 0.01) and delta is something we find out (e.g. 0.087, according to WolframAlpha, in this case)? Does this help you find the limit in the first place?

    This doesn't actually help you find a limit then?
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    (Original post by Zygroth)
    Aah, so we're choosing how close to the function we want to be and we find how far away x can be? So say the function is \dfrac{\sin 2x}{x}, and the limit is 2 as x -> 0. Epsilon is something we choose (e.g. 0.01) and delta is something we find out (e.g. 0.087, according to WolframAlpha, in this case)? Does this help you find the limit in the first place?

    This doesn't actually help you find a limit then?
    Yup. We're just normally finding delta as a function of epsilon to be general.
    Nope, you pretty much need to guess what the limits gonna be(normally not overly tough), and then prove you're right. Sometimes finding the limit is pretty hard though(and it gets worse with differentiation between vector spaces of different dimensions:p:)
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    (Original post by Slumpy)
    Yup. We're just normally finding delta as a function of epsilon to be general.
    Nope, you pretty much need to guess what the limits gonna be(normally not overly tough), and then prove you're right. Sometimes finding the limit is pretty hard though(and it gets worse with differentiation between vector spaces of different dimensions:p:)
    So it's just me not being able to do it rather than this being what I needed to know then :p: How would you guess what the limit is in the first place though? For sin(2x)/x I don't think there's any way I could've gotten 2
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    (Original post by Zygroth)
    So it's just me not being able to do it rather than this being what I needed to know then :p: How would you guess what the limit is in the first place though? For sin(2x)/x I don't think there's any way I could've gotten 2
    Well, I would be surprised if many cases turned up for your course where it wasn't fairly clear, but there might be a few. Quite often limits will be zero. Besides that, I'm not sure really, analysis is not a strong point of mine:p:
    Well, in that case, you could recall that the limit of sin(x)/x-->1 as x-->0, so with a sin(2x) it'll probably be twice that(on the logic sinx~x, so sin2x~2x, for small x.)
    Alternatively, sin2x=2sinxcosx, near 0 cosx is almost 1, and sinx/x tends to 1, so 2 would be a good guess to try.
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    Or, remember that
     \lim_{h \to 0} \frac{f(h)-f(0)}{h}=f'(0)
    Let  f(x)=\sin(2x) . Consequently
     \lim_{h \to 0} \frac{\sin(2h)-\sin(0)}{h}=2\cos(0)=2
    If you see something on the top with a denominator of x and you want to find the limit as x tends to 0 it's quite often just a derivative.
 
 
 
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