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    When you're trying to show that a function is continuous at a point x, and you use the definition.

    Will delta always be min(1, 'and something') ?

    I'm doing a Q and I've nearly finished it: I've got to d(6+d) <=e (this is correct from solutions)

    d= delta
    e= epsilon

    Then in solution it say put delta min(1, e/7)
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    (Original post by Vorsah)
    When you're trying to show that a function is continuous at a point x, and you use the definition.

    Will delta always be min(1, 'and something') ?

    I'm doing a Q and I've nearly finished it: I've got to d(6+d) <=e (this is correct from solutions)

    d= delta
    e= epsilon

    Then in solution it say put delta min(1, e/7)
    Nooooo... Seems like you don't quite understand it yet. It must have been just one example where they used min(1, smth). Delta depends on function and epsilon usually. Minimum ( let alone of 1 and smth ) is nothing universal.
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    (Original post by Vorsah)
    When you're trying to show that a function is continuous at a point x, and you use the definition.

    Will delta always be min(1, 'and something') ?
    No, every question is different. However...

    I'm doing a Q and I've nearly finished it: I've got to d(6+d) <=e (this is correct from solutions)
    There's a really common tactic with questions like this: you want to get to a simple inequality relating d and e, basically simple enough that you can easily "invert it" to say "so if d < {stuff}, then |f(x+d) - f(x)| < e". And conceptually[, you're thinking of e and d as being very small quantities.

    So you want to be able to say "d is small, so 6+d < 7, so d(6+d) < 7d, so if 7d < e, we're good, so we need d < e/7"

    And this is fine, and this is in fact the correct "main" argument - it says what happens when epsilon is small, which is what this is all really about.

    The only problem is, if epsilon is large (in this case, if e > 7) , you find the argument doesn't force d < 1, and then your claim that 6+d > 7 isn't true.

    So to cover this, you need to be a little more careful. Something like:

    As long as d < 1 (*), 6+d < 7, so d(6+d) < 7d ..., so we need d < e/7.
    To ensure (*) holds, we also require d < 1, and so d < min(1, e/7).

    So doing something like this is very common, but it should be a natural part of your argument, not something you "just do without knowing why".
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    (Original post by DFranklin)
    No, every question is different. However...

    There's a really common tactic with questions like this: you want to get to a simple inequality relating d and e, basically simple enough that you can easily "invert it" to say "so if d < {stuff}, then |f(x+d) - f(x)| < e". And conceptually[, you're thinking of e and d as being very small quantities.

    So you want to be able to say "d is small, so 6+d < 7, so d(6+d) < 7d, so if 7d < e, we're good, so we need d < e/7"

    And this is fine, and this is in fact the correct "main" argument - it says what happens when epsilon is small, which is what this is all really about.

    The only problem is, if epsilon is large (in this case, if e > 7) , you find the argument doesn't force d < 1, and then your claim that 6+d > 7 isn't true.

    So to cover this, you need to be a little more careful. Something like:

    As long as d < 1 (*), 6+d < 7, so d(6+d) < 7d ..., so we need d < e/7.
    To ensure (*) holds, we also require d < 1, and so d < min(1, e/7).

    So doing something like this is very common, but it should be a natural part of your argument, not something you "just do without knowing why".
    My friend did this Q a different way compared to the solutions. He assumed d<1 at the start and then manipulated the inequalities and got the same answer. Is his approach where he assumed d<1 at the start correct?

    E.g if we had |x-3|<d, assume d<1 then |x-3|<1 and from here he carried on.
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    (Original post by Vorsah)
    My friend did this Q a different way compared to the solutions. He assumed d<1 at the start and then manipulated the inequalities and got the same answer. Is his approach where he assumed d<1 at the start correct?
    Yes this is fine; strictly speaking you need to make sure your final choice for d satsifies any assumptions you made along the way (i.e. at the end you'd still have a "d < min(1, ...)" type condition to ensure that d < 1).

    As I said in the previous post, the "main" argument is about what happens when e, d are very small, and so as time goes on (and you've proved you can do it "properly") you may find your lecturers/examiners find it acceptable not to worry about the" extra" conditions (such as "d < 1") - a simple "for d sufficiently small" may suffice.

    (e.g. for your problem you might write something like:

    "for d sufficiently small, 6+d < 7 and so d(6+d) < 7d, and so d < e/7 works").

    But from what you've posted it's clear your examiners currently expect those extra conditions to be included.
 
 
 
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