Let $ f:(-1, \infty) \rightarrow \mathbb{R} $ defined by $ f(x)=\frac{x}{\sqrt{1+x}} $ for all $ x>-1 $.
i) Determine the second order Taylor polynomial with evolution point $ a=0 $ of $ f $, which we denote hereafter by $ T^{(2)} f(x ; 0) $.
ii) Let $ \varepsilon>0 $ be arbitrary. Determine a $ \delta>0 $ such that for all $ x \in(0, \delta) $ the inequality $ \left|f(x)-T^{(2)} f(x ; 0)\right|<\varepsilon $ holds.
Problem/approach: For i.) I found the Taylor polynomial and that would be $T^{(2)}f(x;0)=x-\frac{1}{2}x^2$.
But I don't understand ii.) what delta means in this context and then how to find this. Can someone explain to me what is actually being looked for and give me a basic idea on how to approach this?
You have, by Lagrange's formula for the remainder of a Taylor polynomial:$$f(x)-\left(x-\frac12x^2\right)=\frac{f'''(\xi)}{6}x^3,$$for some $\xi$ between $0$ and $x$. On the other hand,$$f'''(\xi)=\frac{3(\xi+6)}{8(\xi+1)^{7/2}},$$and therefore\begin{align}\left|f(x)-\left(x-\frac12x^2\right)\right|&=\left|\frac{(\xi+6)}{16(\xi+1)^{7/2}}x^3\right|\\&\leqslant\frac{|(x+6)x^3|}{16}\\&\leqslant\frac{(\delta+6)\delta^3}{16}.\end{align}If $\delta\leqslant1$, then you have$$\frac{(\delta+6)\delta^3}{16}\leqslant\frac7{16}\delta^3<\frac{\delta^3}2.$$You want this to be smaller than $\varepsilon$. So, take $\delta=\max\left\{1,\sqrt[3]{2\varepsilon}\right\}$.