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Step-by-Step Solution
Step 1: Expand $(1 + x)^{\frac{3}{2}}$ up to $x^2$ terms
Using the Binomial expansion for small $x$ (and ignoring higher powers like $x^3$ and beyond):
$$(1 + x)^{\frac{3}{2}} = 1 + \frac{3}{2} x + \frac{\frac{3}{2}\left(\frac{3}{2} - 1\right)}{2!} x^2 + \dots$$
Simplifying the coefficient of $x^2$:
$$\frac{3}{2} \left(\frac{3}{2}-1\right) = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4},$$
and dividing by $2! = 2$ gives
$$\frac{3}{4} \div 2 = \frac{3}{8}.$$
So up to $x^2$ terms,
$$(1 + x)^{\frac{3}{2}} \approx 1 + \frac{3}{2} x + \frac{3}{8} x^2.$$
Step 2: Expand $(1 + \frac{x}{2})^3$ up to $x^2$ terms
Using the Binomial expansion again:
$$(1 + \tfrac{x}{2})^3 = 1 + 3 \left(\tfrac{x}{2}\right) + 3 \left(\tfrac{x}{2}\right)^2 + \dots$$
which simplifies to
$$1 + \tfrac{3x}{2} + \tfrac{3x^2}{4}.$$
Step 3: Expand $(1 - x)^{\frac{1}{2}}$ (denominator) up to $x^1$ terms
For very small $x$,
$$(1 - x)^{\frac{1}{2}} \approx 1 - \tfrac{x}{2}.$$
However, in this particular problem, we will see how it appears when we divide. If more terms are required, we would expand further, but the main simplification typically comes from the numerator’s $x^2$ terms.
Step 4: Consider the full expression
The expression to approximate is:
$$\frac{(1 + x)^{\frac{3}{2}} - (1 + \tfrac{x}{2})^3}{(1 - x)^{\frac{1}{2}}}.$$
Substitute the expansions from Steps 1 and 2 into the numerator:
$$(1 + \tfrac{3}{2} x + \tfrac{3}{8} x^2) - \bigl(1 + \tfrac{3x}{2} + \tfrac{3x^2}{4}\bigr).$$
This simplifies to
$$\left(\tfrac{3}{8} x^2\right) - \left(\tfrac{3x^2}{4}\right) = \tfrac{3}{8} x^2 - \tfrac{3}{4} x^2 = -\tfrac{3}{8} x^2.$$
Hence, the numerator up to the $x^2$ term is approximately $-\tfrac{3}{8}x^2$.
Step 5: Divide by $(1 - x)^{\frac{1}{2}}$
Let us look at the denominator $(1 - x)^{\tfrac{1}{2}}$. For small $x$:
$$(1 - x)^{\tfrac{1}{2}} \approx 1 - \tfrac{x}{2}.$$
When we divide $-\frac{3}{8} x^2$ by approximately $1 - \tfrac{x}{2}$, the leading term remains in $x^2$, and higher powers would be of order $x^3$ (or beyond), which we neglect:
$$\frac{-\tfrac{3}{8} x^2}{1 - \tfrac{x}{2}} \approx -\tfrac{3}{8} x^2 \left(1 + \tfrac{x}{2} + \dots\right) \approx -\tfrac{3}{8} x^2.$$
Therefore, to the required order (ignoring $x^3$ and higher), the expression is:
$$-\tfrac{3}{8}\,x^2.$$
Final Approximation
Therefore, for sufficiently small $x$, the given expression approximates to:
$$\boxed{-\tfrac{3}{8}\,x^2}.$$
