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    Ok I'm kinda struggling with these quetsions any help would be appreciated.

    Warwick Analysis II 2008 Q2b+c

    Prove from the epsilon-delta definition, that continuity of f at c = 0 holds.

    i) f : R -> R given by f(x) = 2x/(x+2) for f(x) < 0 and f(x) = x for x \ge 0

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    I don't know how to do this. I get the feeling any delta I pick will do but I really don't know how to write this up formally.
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    [/latex]


    ii) f : R -> R given by f(x) = 3xsin(1/x)
    if x \ne 0 and f(x) = 0 if x = 0

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    I think I found this the easiest. Let \delta = \frac{\epsilon}{3}. And so if |x| &lt; \delta we see that |3x\sin\frac{1}{x}| \le |3x| &lt; \epsilon and so f is continuous at 0


    Prove from the epsilon-delta definition, that continuity of f at c = 0 does not hold.

    i) f : R -> R given by f(x) = (x-2)/10 if x \ne 0 and f(x) = 0 if x = 0

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    I'm thinking about the discontinuity definition here - i.e. there exists epsilon > 0 such that for all delta > 0 we have |x| < delta but |
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    (x-2)/10| \ge \epsilon. BUT WHAT'S THE EPSILON?!


    ii) f : R -> R given by f(x) = 3sin(1/x)
    if x \ne 0 and f(x) = 0 if x = 0

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    No idea
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    (Original post by Maths Failure)
    Prove from the epsilon-delta definition, that continuity of f at c = 0 holds.

    i) f : R -> R given by f(x) = 2x/(x+2) for f(x) < 0 and f(x) = x for x \ge 0
    |2x/(x+2)|<|2x| when |x|<1

    (Original post by Maths Failure)
    Prove from the epsilon-delta definition, that continuity of f at c = 0 does not hold.

    i) f : R -> R given by f(x) = (x-2)/10 if x \ne 0 and f(x) = 0 if x = 0
    Prove that the limits on both sides is -1/5.

    (Original post by Maths Failure)
    ii) f : R -> R given by f(x) = 3sin(1/x) if x \ne 0 and f(x) = 0 if x = 0
    Assume the limit is 0 and set up a contradiction using the definition.
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    (Original post by Lord of the Flies)
    |2x/(x+2)|<|2x| when |x|<1
    Ok cheers


    (Original post by Lord of the Flies)
    Prove that the limits on both sides is -1/5.
    How does that help with delta-epsilon? By the way, I mistyped it. It should be:

    f : R -> R given by f(x) = (x-2)/10 if x < 0 and f(x) = x if x \ge 0

    (Original post by Lord of the Flies)
    Assume the limit is 0 and set up a contradiction using the definition.
    I tried this but couldn't get anywhere. More help?
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    (Original post by Maths Failure)
    How does that help with delta-epsilon? By the way, I mistyped it. It should be:

    f : R -> R given by f(x) = (x-2)/10 if x < 0 and f(x) = x if x \ge 0
    Use \delta,\epsilon to prove that the limits are different depending on which side we approach 0. By definition the function is discontinuous at 0.

    (Original post by Maths Failure)
    I tried this but couldn't get anywhere. More help?
    Show that for any \delta there will always exist |x_0|&lt;\delta such that |3\sin \frac{1}{x_0}|=3 and hence \epsilon cannot be made arbitrarily small.

    Hint:

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    x=\frac{2}{k\pi}\Rightarrow |3\sin \frac{1}{x}|=3. In other words, show that there is always an integer k such that |\frac{2}{k\pi}|&lt;\delta for any fixed \delta
 
 
 
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