Delta-Epsilon Proofs

  • #1

Homework Statement



Suppose the functions f and g have the following property: for all E > 0 and all x,

if 0 < |x - 2| < sin((E^2)/9) + E, then |f(x) - 2| < E,
if 0 < |x - 2| < E^2, then |g(x) - 4| < E.

For each E > 0, find a d > 0 such that, for all x,

i) if 0 < |x - 2| < d, then |f(x) + g(x) - 6| < E.


Homework Equations



N/A, I think.

The Attempt at a Solution



Well, what I did was look at |f(x) + g(x) - 6| < E. Since I was given |f(x) - 2| < E and |g(x) - 4| < E, the best strategy seemed to be to change d so that it would produce values that would be, for each expression involving f(x) and g(x) would be less than E/2. However, since I don't actually know what f(x) and g(x) are, I'm at a loss as to how to do that.

Spivak's solution (since this problem comes from there, ch. 5 #6), says the same thing ("we need...< E/2") but then says that this means I need:

0 < |x - 2| < min(sin(E^2/36)^2 + E/2, E^2/4) = d

...the logic of which escapes me.
 
Last edited:

Answers and Replies

  • #2
Gib Z
Homework Helper
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You have d as a function of E for both f(x) and g(x). If you replace E by E/2 everywhere, wouldn't you get what you need?

Then, for the final step that Spivak uses, remember that the value of d you get for f(x) will only guarantee that |f(x)-2| < E/2, and the value of d for g(x) will only guarantee that |g(x) -2| < E/2, so to ensure both are < E/2, you have the pick the smaller of the two d's.
 
  • #3
You have d as a function of E for both f(x) and g(x). If you replace E by E/2 everywhere, wouldn't you get what you need?

Then, for the final step that Spivak uses, remember that the value of d you get for f(x) will only guarantee that |f(x)-2| < E/2, and the value of d for g(x) will only guarantee that |g(x) -2| < E/2, so to ensure both are < E/2, you have the pick the smaller of the two d's.
Thanks for the response.

My problem with this is I don't see how replacing E with E/2 (in delta) replaces E with E/2 for...E. In other words, I don't see why bringing x even closer to a necessarily makes E...wait. I think I see what you're saying.

My mistake, I believe, was that I was imagining E as some fixed value, and E/2 as some number that would be closer to l. I think I see why this would not be the case...thanks!
 

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