Question

If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected, then $\frac{\left(1+x\right)^{\frac{3}{2}} - \left(1+ \frac{1}{2}x\right)^{3}}{\left(1-x\right)^{\frac{1}{2}}}$ may be approximated as

• A. $1 - \frac{3}{8} x^2$
• B. $3x + \frac{3}{8} x^2$
• C. $- \frac{3}{8} x^2$
• D. $\frac{x}{2} - \frac{3}{8} x^2$

Answer 1

Answer: (C)

$- \frac{3}{8} x^2$

Answer 2

$\because \, x^3$ and higher powers of x may be neglected $\therefore \frac{\left(1+x\right) \frac{3}{2} -\left(1+\frac{x}{2}\right)^{3}}{\left(1-x^{\frac{1}{2}}\right)}$ $=\left(1-x\right)^{\frac{-1}{2}} \left[\left(1+\frac{3}{2} x + \frac{\frac{3}{2}. \frac{1}{2}}{2!} x^{2}\right) - \left(1+ \frac{3x}{2} + \frac{3.2}{2!} \frac{x^{2}}{4}\right)\right]$ $= \left[1+ \frac{x}{2} + \frac{\frac{1}{2} . \frac{3}{2}}{2!}x^{2}\right] \left[ \frac{-3}{8} x^{2}\right] = \frac{-3}{8} x^{2}$ (as $x^3$ and higher powers of x can be neglected)