Question

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

The correct option is C $-\frac{3}{8}{x}^2$

$\frac{(1+x{)}^{3/2}-(1+x/2{)}^3}{(1-x{)}^{1/2}}=\frac{(1+\frac{3}{2}x+\frac{3}{8}{x}^2+...)-(1+\frac{3}{2}x+\frac{3}{4}x)+...}{(1-\frac{x}{2}-\frac{x^2}{8}+...)}=(-3/8x^2)(1-\frac{x}{2}+\frac{x^2}{8}+....)=-\frac{3}{8}{x}^2(1+\frac{x}{2}-\frac{x^2}{8}+...)=\frac{-3x^2}{8}$