We want to find the limit of the expression $\frac{\sin x - \sin x \cdot \cos x}{x^3}$ as $x$ approaches $0$.

1. Problem Description

We want to find the limit of the expression $\frac{\sin x - \sin x \cdot \cos x}{x^3}$ as $x$ approaches $0$.

2. Solution Steps

First, we can factor out $\sin x$ from the numerator:

$\lim_{x \to 0} \frac{\sin x - \sin x \cdot \cos x}{x^3} = \lim_{x \to 0} \frac{\sin x (1 - \cos x)}{x^3}$.

We know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$. Therefore, we can rewrite the expression as:

$\lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2}$.

We know that $1 - \cos x = 2 \sin^2(\frac{x}{2})$.

So, we can rewrite the expression as:

$\lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{2 \sin^2(\frac{x}{2})}{x^2}$.

We can also write it as:

$\lim_{x \to 0} \frac{\sin x}{x} \cdot 2 \cdot \frac{\sin^2(\frac{x}{2})}{x^2} = \lim_{x \to 0} \frac{\sin x}{x} \cdot 2 \cdot \frac{\sin(\frac{x}{2})}{x} \cdot \frac{\sin(\frac{x}{2})}{x}$.

$\lim_{x \to 0} \frac{\sin x}{x} \cdot 2 \cdot \frac{\sin(\frac{x}{2})}{x} \cdot \frac{\sin(\frac{x}{2})}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot 2 \cdot \frac{\sin(\frac{x}{2})}{\frac{x}{2}} \cdot \frac{1}{2} \cdot \frac{\sin(\frac{x}{2})}{\frac{x}{2}} \cdot \frac{1}{2}$.

$\lim_{x \to 0} \frac{\sin x}{x} \cdot 2 \cdot \frac{1}{4} \cdot \frac{\sin(\frac{x}{2})}{\frac{x}{2}} \cdot \frac{\sin(\frac{x}{2})}{\frac{x}{2}}$.

Since $\lim_{x \to 0} \frac{\sin ax}{ax} = 1$,

$\lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2} = 1 \cdot \frac{2 \cdot \sin^2(\frac{x}{2})}{x^2} = 1 \cdot \lim_{x \to 0} \frac{2 \sin^2(\frac{x}{2})}{x^2} = 2 \cdot \lim_{x \to 0} \frac{\sin^2(\frac{x}{2})}{x^2}$.

$= 2 \cdot \lim_{x \to 0} \left( \frac{\sin(\frac{x}{2})}{x} \right)^2 = 2 \cdot \lim_{x \to 0} \left( \frac{\sin(\frac{x}{2})}{\frac{x}{2}} \cdot \frac{1}{2} \right)^2 = 2 \cdot \left( 1 \cdot \frac{1}{2} \right)^2 = 2 \cdot \frac{1}{4} = \frac{1}{2}$.

Alternatively, we can use L'Hopital's rule.

$\lim_{x \to 0} \frac{\sin x - \sin x \cos x}{x^3}$. Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's rule.

Taking the derivative of the numerator and denominator, we get:

$\lim_{x \to 0} \frac{\cos x - (\cos x \cos x + \sin x (-\sin x))}{3x^2} = \lim_{x \to 0} \frac{\cos x - \cos^2 x + \sin^2 x}{3x^2} = \lim_{x \to 0} \frac{\cos x - \cos^2 x + 1 - \cos^2 x}{3x^2} = \lim_{x \to 0} \frac{\cos x - 2 \cos^2 x + 1}{3x^2}$.

$= \lim_{x \to 0} \frac{(\cos x - 1)(1 - 2 \cos x)}{3x^2}$.

$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$.

$\lim_{x \to 0} \frac{\cos x - 2 \cos^2 x + 1}{3x^2} = \lim_{x \to 0} \frac{(\cos x - 1)(2 \cos x - 1)}{-3x^2} = \lim_{x \to 0} \frac{(1 - \cos x)(2 \cos x - 1)}{3x^2} = \lim_{x \to 0} \frac{1 - \cos x}{x^2} \cdot \frac{2 \cos x - 1}{3} = \frac{1}{2} \cdot \frac{2 \cdot 1 - 1}{3} = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$.

Going back to $\lim_{x \to 0} \frac{\cos x - 2 \cos^2 x + 1}{3x^2}$. Differentiating again,

$\lim_{x \to 0} \frac{-\sin x + 4 \cos x \sin x}{6x} = \lim_{x \to 0} \frac{\sin x (4 \cos x - 1)}{6x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{4 \cos x - 1}{6} = 1 \cdot \frac{4 \cdot 1 - 1}{6} = \frac{3}{6} = \frac{1}{2}$.

3. Final Answer

The final answer is $\frac{1}{2}$