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# Limits help watch

1. 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)
2. (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.
3. (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".
4. (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.
5. (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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Updated: October 25, 2015
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