Someone helps me please. I don't know how to do what my prof asked me to do. He said: "You need an argument explaining how to put together [0,2] and [1,infinitive) and conclude uniformly continuity on [0,infinitive)"
Is it not that those intervals are overlap makes the function uniformly continuous on the whole [0, infinitive)
Please, help

Question

Someone helps me please. I don't know how to do what my prof asked me to do. He said: "You need an argument explaining how to put together [0,2] and [1,infinitive) and conclude uniformly continuity on [0,infinitive)"
Is it not that those intervals are overlap makes the function uniformly continuous on the whole [0, infinitive)
Please, help

loser66 · Answer

attachment

freckles · Answer

I'm on my phone looking at this so I might have to comment later.  That is if I can once I have read your pdf.

loser66 · Answer

It's ok, I do appreciate. Whenever you are comfortable, please leave the comments. :)

loser66 · Answer

@pitamar

loser66 · Answer

Those are my prof's lecture

freckles · Answer

So the first document was what you submitted to your teacher and the red markings are from your teacher?

freckles · Answer

hmm I don't think I could add anymore other than what you added
you know that the [0,2] union [1,inf)=[0,inf)

loser66 · Answer

|dw:1426291117016:dw|
$\sqrt{x_2}-\sqrt{x_1}$ what we consider as $\varepsilon$ , but we can determine $\sqrt{dx}$ right? @pitamar

loser66 · Answer

however, your argument still hold at the derivative of the point near 0 is steepest

loser66 · Answer

or, largest if you like, since $f'(x) =\dfrac{1}{2\sqrt x}$  and as x--> 0 $\lim_{x\rightarrow 0}f'(x) = \infty$

loser66 · Answer

hence any point after 0 has the tangent line less as steep as it is at x =0

loser66 · Answer

then?