To solve the limit $\lim_{x \to 0} \frac{x - \sin x}{x - \tan x}$, we proceed as follows:

---

Step 1: Analyze the behavior of $\sin x$ and $\tan x$ near $x = 0$

• The small-angle approximations for $\sin x$ and $\tan x$ as $x \to 0$ are: $\sin x \approx x - \frac{x^3}{6} + \mathcal{O}(x^5), \quad \tan x \approx x + \frac{x^3}{3} + \mathcal{O}(x^5).$

Step 2: Expand $x - \sin x$ and $x - \tan x$ using approximations

1. For $x - \sin x$: $x - \sin x \approx x - \left( x - \frac{x^3}{6} \right) = \frac{x^3}{6} + \mathcal{O}(x^5).$

2. For $x - \tan x$: $x - \tan x \approx x - \left( x + \frac{x^3}{3} \right) = -\frac{x^3}{3} + \mathcal{O}(x^5).$

Thus, the numerator and denominator behave like:

• Numerator: $x - \sin x \sim \frac{x^3}{6}$,
• Denominator: $x - \tan x \sim -\frac{x^3}{3}$.

---

Step 3: Combine terms into the limit

Substitute the approximations into the original limit: $\lim_{x \to 0} \frac{x - \sin x}{x - \tan x} \sim \lim_{x \to 0} \frac{\frac{x^3}{6}}{-\frac{x^3}{3}}.$

Simplify: $\frac{\frac{x^3}{6}}{-\frac{x^3}{3}} = \frac{1}{6} \cdot \frac{3}{-1} = -\frac{1}{2}.$

---

Final Answer:

$\lim_{x \to 0} \frac{x - \sin x}{x - \tan x} = -\frac{1}{2}.$

---

Do you need further clarification or additional steps? Let me know!

Related Questions:

1. What are small-angle approximations for $\cos x$ and higher-order terms?
2. How would the limit change if the denominator were $x - \cos x$?
3. Explain why higher-order terms are ignored when calculating limits.
4. How do we derive Taylor expansions for $\sin x$ and $\tan x$?
5. What happens to the limit if we change $x \to \infty$ instead of $x \to 0$?

Tip:

When solving limits involving trigonometric functions near $x = 0$, always consider using Taylor series expansions for more accurate results.