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    Can someone check my working is correct here please? I just want to make sure I understand things correctly here.
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    Yep, you got the correct bounds. This time you've shown that your delta will work (like what james22 said in the last thread) albeit in a rather hasty way at the bottom, but in a formal proof you'd have to be more explicit.

    But yes, you've got it.
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    (Original post by 0x2a)
    Yep, you got the correct bounds. This time you've shown that your delta will work (like what james22 said in the last thread) albeit in a rather hasty way at the bottom, but in a formal proof you'd have to be more explicit.

    But yes, you've got it.
    Could you make an suggestions for the improvement of my answer please? as understand it, I just proved that the function is continuous at 0, because i was able to find a delta, right?

    Thanks a lot for the help
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    (Original post by pineapplechemist)
    Could you make an suggestions for the improvement of my answer please? as understand it, I just proved that the function is continuous at 0, because i was able to find a delta, right?

    Thanks a lot for the help
    You haven't necessarily proved that it is continuous, because for a function to be continuous at a point x_0 there must exist a \delta for all \varepsilon, such that when |x - x_0| < \delta then |f(x) - f(x_0)| < \varepsilon, but the thing is, once you have a concrete example like 1/100 in your case, in most circumstances it is quite easy to extend this to any general \varepsilon.

    Other than that, the thing I was nitpicking at was to write a formal proof at the bottom.

    i.e. Choose any \varepsilon, let \delta = \varepsilon, and let |x - 0| = |x| < \delta = \varepsilon.

    Then \displaystyle |f(x) - f(0)| = \left|x\sin\dfrac{1}{x} - 0 \right| = \left|x\sin\dfrac{1}{x}\right| = |x|\left|\sin\dfrac{1}{x}\right| \leq |x| < \delta = \varepsilon, thus |f(x) - f(0)| < \varepsilon, whenever |x - 0| < \delta.
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    (Original post by 0x2a)
    You haven't necessarily proved that it is continuous, because for a function to be continuous at a point x_0 there must exist a \delta for all \varepsilon, such that when |x - x_0| < \delta then |f(x) - f(x_0)| < \varepsilon, but the thing is, once you have a concrete example like 1/100 in your case, in most circumstances it is quite easy to extend this to any general \varepsilon.

    Other than that, the thing I was nitpicking at was to write a formal proof at the bottom.

    i.e. Choose any \varepsilon, let \delta = \varepsilon, and let |x - 0| = |x| < \delta = \varepsilon.

    Then \displaystyle |f(x) - f(0)| = \left|x\sin\dfrac{1}{x} - 0 \right| = \left|x\sin\dfrac{1}{x}\right| = |x|\left|\sin\dfrac{1}{x}\right| \leq |x| < \delta = \varepsilon, thus |f(x) - f(0)| < \varepsilon, whenever |x - 0| < \delta.
    Thanks so much!

    So because we can choose epsilon=delta the function is continuous at 0?
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    (Original post by pineapplechemist)
    Thanks so much!

    So because we can choose epsilon=delta the function is continuous at 0?
    Yes, in this case the delta we need is easy because we can just set it to epsilon.

    Np
 
 
 
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