Proof using epsilons and deltas.
Can anyone please show me how to prove the problem below using epsilons and deltas. Thanks a lot...I'm so stuck...
I got through this step: /(5x5)/(3+x)/ < Epsilon, and don't know where to go from there. Plz help me out, thanx very much for ur valuable time. ***Prove, using epsilons and deltas, that the limit as x approaches (1) of the function f(x) = 10/(3 + x) is 5. (Can you think of a pedagogical reason for that change?) Hint: you'll have to fiddle around and you'll probably have to finally choose your delta to be the smaller of two numbers.**** 
You have to find a d>0 such that if x+1<d then f(x)5<e
You have been given an epsilon, you must show that there exists some d>0 corresponding to this epsilon (and for all epsilons). So, you must find some sort of relation between e and d. Given this value of e, where must x lie? If you factor 5 from 5x5 you get 5(x+1)=5x+1. Try solving for x+1. Bear in mind that you want x+1<d. 
A Steven Privatera noted, you want f(x)5< epsilon and you have already reduced that to (5x5)/(3+x)< epsilon which is, of course, the same as 5x(1)/3+x< epsilon. That is the same as
x(1)< (epsilon/5)3+x. The problem is that you need a constant on the right, not something that depends on x! Okay: you know you want x close to 1. Let's for the moment assume that x+1< 1 (just because 1 happens to be easy to work with). 1< x+1< 1 so 1+ 2<x+1+ 2< 1+2 or 1< x+3< 3. Since 1< x+3, (epsilon/5)< (epsilon/5)3+x. (Notice the direction of the inequality.) So if we take x(1) less than the SMALLER of epsilon/5 AND 1, we have both x+1< epsilon/5 and epsilon/5< (epsilon/5)3+x so x+1< (epsilon/5x+3 and you can work backwards to get 5x+1/x+3= 10/(3 + x)5< epsilon which is what you want. 
I think that you might want to show that the change in f(x) approaches zero as the change in x approaches zero. That way, the function wont do anything wild or crazy around the point at which we are calculating the limit, and will approach its calculated value in a smooth manner. To me that shows that the limit of the function as x approaches any value is the function of the value itself...
show that f(x+dx)  f(x) [the change in f(x)] is zero for dx = 0: f(x) = 10 / (3  x) f(x+dx) = 10 / (3  (x+dx)) f(x+dx)  f(x) = (10 / (3  (x+dx)))  (10 / (3  x)) (make common denominator): = ((10 * (3  x)) / (3  (x+dx)) * (3  x))  (10 * (3  (x+dx)) / (3  x) * (3  (x+dx)) = ((10 * (3  x))  (10 * (3  (x+dx)) / ((3  x) * (3  (x+dx))) (extend all the terms): =(30  10x  30 + 10x + 10dx) / (9  3x 3x  3dx + x*x  xdx) =10dx / (9  6x 3dx + x*x  xdx) (add (3x)dx to top & bottom): =10dx / (9  6x + x*x  (3x)dx) =(10dx + (3x)dx) / (9  6x + x*x) =((10 + (3x))dx) / (9  6x + x*x). Now, since dx is a factor in the numerator, and is eliminated from the denominator, I say that proves this expression, which is the change in f(x) corresponding to a change in x, goes to zero as dx goes to zero. I hope this helps. I'm not sure what an epsilon is; I think the delta is my dx. Eric 
the request was to use the epsilon delta definition of limit and not infinitesimals.

Hello, gigi9!
I have an answer. (But am I oversimplifying the problem?) We have: 10/(3 + x)  5 < e or: (5x  5)(3 + x) < e and: x  (1) < d or: x + 1 < d The epsilon statement becomes: (5)(x + 1)/(3 + x) < e 5x + 1/3 + x < e 5x + 1/3 + x < e If x is "close" to 1, then (3 + x) is "close" to +2. Then we have: 5x + 1 < e/(3 + x) < e or: x + 1 < e/5 I believe we can use: d = e/5 
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