iF $\displaystyle g(x)=\sqrt{x}$ for $\displaystyle x\in [0,1]$ , how would you show that there doesn't exist a constant $\displaystyle K$ such that $\displaystyle |g(x)|\leq K|x|$ , for $\displaystyle x\in [0,1]$
$\displaystyle |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 $\displaystyle \lim_{x\to 0} \frac{1}{\sqrt{x}} \leq K $, but $\displaystyle \lim_{x\to 0} \frac{1}{\sqrt{x}} = \infty $.
Therefore $\displaystyle K $ can't exist for all $\displaystyle x $ in this domain.