Problem source

Use the binomial expansion to obtain
  a polynomial of degree $2$ which is a good approximation
 to $\sqrt{1-x}$ when $x$ is small.  
 \begin{questionparts}
 \item
 By taking  $x=1/100$, show that $\sqrt{11}\approx79599/24000$,
 and estimate, correct to 1 significant figure, 
 the error in this approximation. (You may assume that the error is given approximately by the 
 first neglected term in the binomial expansion.)

 \item
 Find a rational number which approximates  $\sqrt{1111}$ with an error
 of about  $2 \times {10}^{-12}$.
 \end{questionparts}

Solution source

\begin{align*}
&& \sqrt{1-x} &= (1-x)^{\frac12} \\
&&&= 1 -\frac12x+\frac{\frac12 \cdot \left (-\frac12 \right)}{2!}x^2 + \frac{\frac12 \cdot \left (-\frac12 \right) \cdot \left (-\frac32 \right)}{3!} x^3\cdots \\
&&&\approx 1-\frac12x - \frac18x^2
\end{align*}

\begin{questionparts}
\item $\,$

\begin{align*}
&& \frac{3\sqrt{11}}{10} &= \sqrt{1-1/100} \\
&&&\approx 1 - \frac{1}2 \frac{1}{100} - \frac{1}{8} \frac{1}{100^2} \\
&&&= \frac{80000-400-1}{80000} \\
&&&= \frac{79599}{80000}\\
\Rightarrow && \sqrt{11} &\approx \frac{79599}{24000} \\
\\
&&\text{error} &\approx \frac{1}{16} \frac{10}3 \frac{1}{100^3} \\
&&&= \frac{1}{48} 10^{-5} \\
&&&\approx 2 \times 10^{-7}
\end{align*} 

\item Taking $x = 1/10^4$ we have

\begin{align*}
&& \frac{3 \sqrt{1111}}{100} &= \sqrt{1-1/10^4} \\
&&&\approx 1 - \frac12 \frac1{10^4} - \frac18 \frac{1}{10^8} \\
&&&= \frac{799959999}{800000000} \\
\Rightarrow && \sqrt{1111} & \approx \frac{266653333}{8000000}  \\
\\
&& \text{error} &\approx \frac{100}{3} \frac{1}{16} \frac{1}{10^{12}} \\
&&&= \frac{1}{48} \frac{1}{10^{10}} \\
&&&\approx 2 \times 10^{-12}
\end{align*}

\end{questionparts}