Proving Uniform Continuity of f(x): Let x in [Infinity, 0)

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Homework Help Overview

The problem involves proving that the function f(x) = (x^2)/(1+x) is uniformly continuous on the interval [infinity, 0). Participants are discussing the definition and properties of uniform continuity in the context of this function.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to simplify the function using long division. Some participants question the definition of the domain and the meaning of "uniformly continuous." Others suggest finding a delta that does not depend on specific points in the domain.

Discussion Status

The discussion is exploring various interpretations of the problem, particularly regarding the domain of the function and the definition of uniform continuity. Some participants have provided guidance on how to express the uniform continuity condition mathematically, while others are clarifying the assumptions made in the problem statement.

Contextual Notes

There is confusion regarding the domain notation "[infinity, 0)" which some participants suggest should be corrected to "[0, infinity)". The discussion also highlights the importance of ensuring that delta is independent of specific points in the domain.

minyo
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Homework Statement



let f(x)= (x^2)/(1+x) for all x in [ifinity, 0) proof that f(x) is uniformly continuous. can anyone help me with this problem

Homework Equations



using the definition of a uniform continuous function

The Attempt at a Solution



i did long division to simplify the problem and got (x-1) - 1/(x+1)
 
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What is the definition of "uniformly continuous" and how have you tried to use it?
 
Find a delta that doesn't depend on any specific point in the domain of f.
 
for u and v in D,

|f(u)-f(v)|<epsilon if |u-v|<delta
 
Do you mean the domain to be [0, infinity) or (0, infinity)? Since "infinity" isn't a real number, "[infinity, 0)" doesn't make sense. I suspect it was supposed to be [0, infinity).

"for u and v in D,

|f(u)-f(v)|<epsilon if |u-v|<delta"

is much too "shorthand" to be useful here. f is uniformly continuous in D if f(x) is defined for all x in d and given epsilon> 0, there exist delta> 0 such that if |u-v|< delta, then |f(u)- f(v)|< epsilon.

The crucial point is, as JG89 said, the delta depends only on epsilon, not u or v.

|f(u)- f(v)|= |u^2/(1+u)- v^2/(1+v)|= |(u^2(1+v)- v^2(1+u))/(1+ u+ v+ uv)|
= |u^2- v^2+ uv(u- v))/(1+ u+ v+ uv)|= |(u+v- uv)/(1+u+v+uv)||u-v|.

If you can find an upper bound for that first fraction, you are home free.
 

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