Question

If $x$ be so small that its square and higher powers may be neglected, find the value of :
$\sqrt{4-x}\cdot {(3-\frac{x}{2})}^{-1}$

Answer 1

$y=(4-x{)}^{0.5}.(3-\frac{x}{2}{)}^{-1}=\frac{2}{3}(1-\frac{x}{4}{)}^{\frac{1}{2}}.(1-\frac{x}{6}{)}^{-1}$
$=\frac{2}{3}[1-\frac{1}{2}.\frac{x}{4}-\frac{1}{4}.\frac{x^2}{16.2!}.......][1+\frac{x}{6}+2\times \frac{x^2}{36.2!}.......]$
since higher powers are neglected , we can neglect power 2 terms as well
$=\frac{2}{3}[1+\frac{x}{6}-\frac{x}{8}]$
$=\frac{2}{3}[1+\frac{x}{24}]$