Evaluating a trigonometric limit via known building-block limits

Start by combining $\tan(x) - \sin(x)$ over a common denominator: $\tan(x) - \sin(x) = \dfrac{\sin(x)}{\cos(x)} - \sin(x) = \dfrac{\sin(x)(1-\cos(x))}{\cos(x)}$. Dividing by $x^3$ gives $h(x) = \dfrac{\sin(x)}{x} \cdot \dfrac{1}{\cos(x)} \cdot \dfrac{1-\cos(x)}{x^2}$, which splits the original expression into three factors each with a known limit as $x \to 0$.

Since each factor has a limit as $x \to 0$ ($\frac{\sin x}{x} \to 1$, $\frac{1}{\cos x} \to 1$, and $\frac{1-\cos x}{x^2} \to \frac12$, the last being a standard consequence of the Squeeze Theorem), the product rule for limits applies. Multiplying these limits together: $1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}$. Therefore $\lim_{x\to 0} \frac{\tan(x)-\sin(x)}{x^3} = \frac{1}{2}$.

The answer is one. Since cosine of zero equals one, the factor one over cosine of x approaches one over one, which is one, as x approaches zero. This is the middle factor in the decomposition from part a, and it is the reason the overall product simplifies to just the limit of the third factor, one half.