Real Analysis: Continuity and Uniform Continuity

[danielkyulee]

Question: Show that f(x)= (x^2)/((x^2)+1) is continuous on [0,infinity). Is it uniformly continuous?


My attempt: So I know that continuity is defined as
"given any Epsilon, and for all x contained in A, there exists delta >0 such that if y is contained in A and abs(y-x)<delta, then abs(f(x)-f(y))<Epsilon.

So i tried expanding the function, but still can not find the values for delta that make this continuous on [0,infinity). Any ideas?

Also, it is NOT uniformly continuous correct?

Thanks!

 

[lanedance]

well its definitely continuous on [0,inf) - can you show that?

now for the "uniform continuity part", not as the actual proof, but I would consider the derivative - it should give you an idea of how to approach the problem

loosely speaking if a function is uniformly continuous it shouldn't have a derivative that misbehaves

consider
|\frac{f(y)-f(x)}{y-x}|
in the limit y goes to x, this becomes the derivative...

 

[danielkyulee]

Hey, so one of the problem I had was that I could not find a way to show it was continuous. I tried to show it using the delta-epsilon definition of continuity, but couldn't figure out a way to find delta.

 

[lanedance]

ok so where did you get stuck?

 

[danielkyulee]

So i did absolute(f(x)-f(y))= abs( ((x^2)/((x^2)+1)) - ((y^2)/((y^2)+1)) ). I simplified this to get abs( ((x^2)-(y^2))/((x^2)+1)((y^2)+1)) ).

Now I tried to set it up so that abs ( x-y ) is less than 1 (which would be one of the minimum components of delta) so that I can try to abs( x-y ) less than something with respect to epsilon. However, I am stuck on how I can use <1 to find the deltas that make abs( f(x)-f(y) ) < E for any given E.

 

[SteveL27]

Question: Show that f(x)= (x^2)/((x^2)+1) is continuous on [0,infinity). Is it uniformly continuous?


My attempt: So I know that continuity is defined as
"given any Epsilon, and for all x contained in A, there exists delta >0 such that if y is contained in A and abs(y-x)<delta, then abs(f(x)-f(y))<Epsilon.

So i tried expanding the function, but still can not find the values for delta that make this continuous on [0,infinity). Any ideas?

Also, it is NOT uniformly continuous correct?

Thanks!

I believe you have your definitions backwards. A function is continuous at a point x if given epsilon there is a delta. If you pick some different point y, then given the same epsilon, you may need a different delta.

If the same delta works for a given epsilon regardless of x, that is uniform continuity.

What you wrote above is the definition of uniform continuity ... given epsilon, for any x there's a delta. The def of continuity is that given x, for any epsilon there is a delta that depends on x.