<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] expanding about $x=0$
$$ \begin{equation} 
  \begin{aligned}
    P_n(x) = f(0)
    + \frac{f'(0)}{1!}(x - 0)
    + \frac{f''(a)}{2!}(x - 0)^2
    & + \ldots
    + \frac{f^{(n)}(0)}{n!}(x - 0)^n \\
  \end{aligned}
\end{equation}\\
$$
$$ \begin{equation} 
  \begin{aligned}
    P_n(x) = 1
    + \frac{1}{2}x     + \frac{-\frac14}{2!}x^2
    & + \ldots
      \end{aligned}
\end{equation}\\
$$so w..r.t.$|x| < 0.01$
$$\begin{equation}
  \begin{aligned}
    R_n(x) & = \frac{- \frac{1}{4(1 + \xi)^{+3/2}} }{2!}
  \end{aligned}
\end{equation}\leq \frac{- \frac{1}{4} (1 + (-0.01))^{-3/2}}{2!}=\frac{- \frac{1}{4} (0.99)^{-3/2}}{2!}$$
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