Is \frac{1}{\sqrt{x}} uniformly continuous in (0,1)?

[nhrock3]

i need to prove that $$\frac{1}{\sqrt{x}}$$ is not uniformly continues in (0,1)







for epsilon=0.5



$$|\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}|=|$$$$]\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}\frac{\sqrt{y}+\sqrt{x}}{\sqrt {y}+\sqrt{x}}|$$$$=|\frac{y-x}{(\sqrt{y}-\sqrt{x})\sqrt{xy}}|$$







i need to prove that the above exprseesion bigger then 0.5







but i don't know what x and y to choose



?

 

[Robert1986]

So, for each delta, you need to find an x and y such that |x-y| is smaller then delta but f(x)-f(y) is less than 1/2, as you have stated. But, realize two things:

1) So long as |x-y| is smaller than delta, you can let |x-y| be anything you want (think about why this is true.) For example, if delta is bigger than zero, you can choose x and y such that (I'm going to write "d" for delta here) d/4 < |x-y| < d. In other words, you aren't proving that it is uniformly continuous, you are proving that it isn't (so think of the negation of the definition of uniform continuity.)

2) multiply $$\frac{1}{\sqrt{y}}-\frac{1}{\sqrt{x}}$$ by $$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}$$ and then use the fact that |x-y| is bigger than something (see the above.)

Does this make sense?