How to Prove a Function is Uniformly Continuous

How to Prove a Function is Uniformly Continuous. This is a proof that f(x) = 1/(1 + x^2) is uniformly continuous on R. I hope this video helps.

[lemyul]

it will probably take me years before this whole analysis course make sense

[smoosq]

I just wanna say thank you, your videos not only explain everything in detail but also saved me a huge amount of time.

[tannerboos]

There is a problem with the yellow proof at the end. Consider a=-5. -5<=1 but |a| is not less than or equal to 1. Not hard to fix mind you, build your cases as |a|<=1 and |a|>1. The rest follows from there in the same way.

[bengper]

The best 4 minutes I ever spent (2x speed works like a dream)

[kpopaspirations]

How the heck did you just arbitrarily drop abs(1+ y^2) from the bottom of the fraction?

[KevinWeatherwalks]

Since 1/|1+x^2| is bounded above by 1 and |x|/|1+x^2| is bounded above by 1/2 (and similarly for y), will the proof still hold if we had chosen delta = epsilon?

[Fragadagalops]

you said that we could just "get rid" of the |1+x^2| and |1+y^2| in denominators but you didn't explain it.

[AsaNole]

Why is the last inequality "less than or equal to 1" instead of strictly less than 1? What I mean is, does there exists a number a such that (|a| / |1+a^2| )= 1?

[CREricNoeJimenezValerio]

Love your proves! Keep it on dude, you are saving lot of asses around the world

[potatojam]

So, if I were cluelessly doing this, I could just choose 𝜺 = 𝛿 and in the end see that the |f(x)-f(y)|<2*𝜺 and create an 𝜺1 such that 𝜺1=2𝛿 and (1/2)*𝜺1=𝛿 as you had it from beginning and then just go back and pretend that I knew it all along and fool my professor into thinking I know what I'm doing?

[javieralonso]

how do you know that the delta value is epsilon/2?

[panos.kardatos]

Awesome video, I really liked the way you make the proof at the end of it about 1+1.

[viiarush]

Nice, you could've also shown x/1+x^2 as a sin function and hence <=1.

[matthutchings]

In the proof at the end, shouldn't you be considering the cases where |a| <= 1 and |a| > 1? The way you did it means that if a < -1, then |a| > 1 and you can't use the first inequality in case 1.

[HXED_OwO]

mr math sorcerer, at 7:30 can we prove this segment by proving the bottom part of the fraction is bigger or equal to the top part?
what i mean by this is by plugging in any number for 'a' we get a result smaller
or equal to one

[mayankjangid]

Thanks professor for your videos. They help us a lot. With due respect, your proof for the lemma used is wrong(for the case when a<=1) for which -2 can be given as a counter example. Thanks!

[Mycrosss]

Um if you decided to set a <= 1, you have to put it in the a^2 too meaning you'd get 1/2, but it didn't hurt you here since 1/2 < 1 and you can choose to continue like it was a 1

[ArunDwivedi]

There is A question in my mind as u select epsilon is greater then 0 I agree but by which logic u said that delta will equal to epsilon/2

[omatseyeugen]

any examples of determining if a function is continuous and or discontinuous?

[eggy]

immediately frustrated because you figured out delta beforehand and was just like ok there you go. now let's do more redundant stuff to see if it's actually what I put down though I already told you it is.