(1)Find an explicit value of $\varepsilon >0$ such that for every $x\in \left [ 0,1 \right ] $, there always is $$\left | \sqrt[]{x} -\sqrt[]{x+\varepsilon} \right | < \frac{1}{200}$$
(2)Find an explicit integer $N$ such that there exists a polynomial $P$ of degree at most $N$ such that for every $x\in \left [ 0,1 \right ] $, there always is $$\left | \sqrt[]{x} -P\left ( x \right ) \right | < \frac{1}{100}$$
First part of the question, can I do it this way?$$\left | \sqrt[]{x}-\sqrt[]{x+\varepsilon } \right | =\left | \frac{x-\left ( x+\varepsilon \right )}{ \sqrt[]{x}+\sqrt[]{x+\varepsilon } } \right |\le \left |\frac{x-\left ( x+\varepsilon \right )}{2\sqrt[]{x} } \right |=\frac{\varepsilon }{2\sqrt[]{x} } $$ next, take$$ \frac{\varepsilon }{2\sqrt[]{x} }=\frac{1}{200} $$ therefore, as long as $\varepsilon=\frac{\sqrt{x}}{100}$, the inequality of (1) hold?(honestly, I feel weird.)
And about second part of the question, the only thing I know is that I should use the expansion of $\sqrt{x+\varepsilon}\space$ in power series of $(x-1)$. Could someone give me some example how I do it?
For the first part we have: $\left|\sqrt{x} - \sqrt{x+\epsilon}\right|= \dfrac{\epsilon}{\sqrt{x}+\sqrt{x+\epsilon}}\le \dfrac{\epsilon}{\sqrt{\epsilon}}=\sqrt{\epsilon} < \dfrac{1}{200}\implies \epsilon < \dfrac{1}{40,000}$. Thus you can take for example $\epsilon = \dfrac{1}{50,000}$ and this epsilon does it for you.