Uniform Continuity (Example 2): Challenging Example

[@김정인-b8v]

Thank you for helping my study! You’ve got great talent at explaining something difficult in easier words. Thank you☺️

[@drpeyam]

Thank you so much!!!

[@Nand0san35]

Love your style! thank u for showing maths in such a beatifull way.

[@bigwit8875]

i finally get it !thank you!

[@pilara3262]

Thank you! This vid is so helpful!

[@sudandevkota7268]

Thank you very much sir. Much appreciated.

[@arturcostasteiner9735]

We could ad well apply the following theorem: if f:I --> R is differentiable on the interval I and sup {|f'(x)| : x is in I} = s in R, then f is Lipschitz with constant s. In our case, f(x) = 1/x^2, I = (2, oo) , f'(x) = -2/x^3, so that s = 2/8 = 1/4. Hence, for all x and y in (2, oo), |1/x^2 - 1/y^2| < |x - y|/4. Here, strict inequality. This theorem is a simple consequence of the Mean Value Theorem. Its converse is also true. It frequently provides an easy way to show a function is Lipschitz, so uniformly continuous, on an interval. For example, it readily shows f(x) = e^(-x) is u. c. on [0, oo).

[@drpeyam]

I prove that later in the playlist, check it out

[@iabervon]

It's kind of fun to find delta by finding 1/sup{|f'|}, and then prove that it works using algebra you hadn't already done. Also neat to see something come up that's similar to 2/x^3 but split up.

[@MrCigarro50]

Muchas Gracias por su video, realmente se le agradece que nos enseñe todas estas cosas de una forma tan amena. En honor a la verdad, sus presentaciones son muy claras y hasta divertidas.

[@drpeyam]

Muchísimas gracias, lo aprecio mucho!!!

[@raiturner2132]

Really helpful, thank you 😊

[@drpeyam]

Glad it was helpful!

[@dimitnone6795]

ε - δ proofs made easy! Thanks!

[@adolfocanahuire4960]

GRACIAS POR LA EXPOSICIÓN.

[@existentialrap521]

on crip good stuff

[@رشاداليامي]

Thanks doctor, it’s perfect video but i have question when i take one value of delta and take delta maximum or minimum while prove using “ epsilon - delta”

[@رشاداليامي]

Sometimes, take delta = constant epsilon and take delta = min ... sorry we’re not take delta = max that’s wrong

[@ZipplyZane]

It seems to me that you only use the min function for Delta when there's no other way to simplify. In this case, since you have a minimum value for both x and y, there is no need to define an upper bound for delta.

[@xinpingdonohoe3978]

They say that √x is uniformly continuous I think, but I thought one criterion is that the limit at ∞ had to exist and be finite for uniform continuity. Have I misunderstood something?

[@drpeyam]

No the limit at infinity doesn’t have to exist, f(x) = x is uniformly continuous on R

[@xinpingdonohoe3978]

@@drpeyam okay, thanks