<s>[INST] Estimate the error that results when $\sqrt{1 + x}$ is replaced by $1 + \frac{1}{2}x$ if $|x| < 0.01$ Question
Estimate the error that results when $\sqrt{1 + x}$ is replaced by $1 + \frac{1}{2}x$ if $|x| < 0.01$ 
Definition
Taylors formula is $f(x) = P_n(x) + R_n(x)$ where $P_n(x)$ is
\begin{equation} 
  \begin{aligned}
    P_n(x) = f(a)
    + \frac{f'(a)}{1!}(x - a)
    + \frac{f''(a)}{2!}(x - a)^2
    & + \ldots
    + \frac{f^{(n)}(a)}{n!}(x - a)^n \\
  \end{aligned}
\end{equation}
And $R_n (x) $ is  (\emph{where $\xi$ is between $a$ and $x$ })
\begin{equation}
  \begin{aligned}
    R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x - a)^{(n + 1)}
  \end{aligned}
\end{equation}

Working
I'm not sure how to go about this, would I say that this is a first order
approximation as
\begin{equation}
  \begin{aligned}
    P(x) & = 1 - \frac{1}{2}x \\
    P'(x) &=  - \frac{1}{2} \\
    P''(x) &= 0
  \end{aligned}
\end{equation}
Then the remainder term would be 
\begin{equation}
  \begin{aligned}
    R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x - a)^{(n + 1)}
  \end{aligned}
\end{equation}
Where $n + 1 = 2$. For $f(x) = \sqrt{1 + x}$ this would be 
\begin{equation}
  \begin{aligned}
    R_n(x) & = \frac{- \frac{1}{4} (1 + \xi)^{-3/2}}{(3)!}
  \end{aligned}
\end{equation}
And $\xi $ is between $-0.01$ and $0.01$ 
This would give the maximum error as 
\begin{equation}
  \begin{aligned}
    R_n(x) & = \frac{- \frac{1}{4} (1 \pm 0.01 )^{-3/2}}{(3)!} \approx -0.0423
  \end{aligned}
\end{equation}
The error is greatest when $\xi = -0.01$.
 [/INST] You have that
$$f(x)=f(0)+\dfrac{f'(0)}{1!}x+R_1(x).$$ Since $f(x)=\sqrt{1+x}$ it is $f(0)=1$ and $f'(0)=\dfrac 12.$ Thus 
$$\sqrt{1+x}=1+\dfrac{1}{2}x+R_1(x).$$
Now, $$R_1(x)=\dfrac{f''(\xi)}{2!}x^2.$$ Since $$f''(\xi)=-\dfrac14 (1+\xi)^{-3/2}$$ we get
$$\sqrt{1+x}-\left(1+\dfrac{1}{2}x\right)=-\dfrac18 (1+\xi)^{-3/2}x^2.$$ Now, if $|x|\le 0.01$ we have to get a bound for the RHT. We have that 
$$\dfrac18 (1+\xi)^{-3/2}x^2\le \dfrac{0.01^2}{8},$$ where we have used that $$(1+\xi)^{-3/2}\le 1.$$
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