# Math Help - More uniform continuity!!

1. ## More uniform continuity!!

iF $g(x)=\sqrt{x}$ for $x\in [0,1]$ , how would you show that there doesn't exist a constant $K$ such that $|g(x)|\leq K|x|$ , for $x\in [0,1]$

2. Originally Posted by CrazyCat87
iF $g(x)=\sqrt{x}$ for $x\in [0,1]$ , how would you show that there doesn't exist a constant $K$ such that $|g(x)|\leq K|x|$ , for $x\in [0,1]$
$|g(x)|\leq K|x| \implies \sqrt{x} \leq Kx \implies \frac{\sqrt{x}}{x} \leq K \implies \frac{1}{\sqrt{x}} \leq K \; \forall x\in [0,1]$.

Hence $\lim_{x\to 0} \frac{1}{\sqrt{x}} \leq K$, but $\lim_{x\to 0} \frac{1}{\sqrt{x}} = \infty$.

Therefore $K$ can't exist for all $x$ in this domain.

3. Could you conclude this is a Lipschitz function?

4. no it's not Lipschitz.

5. why??