Evaluate `lim_(x to 0) (tan x - "sin" x)/(x^(3))`

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} \), we can follow these steps:

### Step 1: Rewrite the Limit
We start with the limit:
\[ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} \]
We know that \( \tan x = \frac{\sin x}{\cos x} \), so we can rewrite the expression:
\[ \lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \]

### Step 2: Combine the Terms
Factor out \( \sin x \):
\[ \lim_{x \to 0} \frac{\sin x \left(\frac{1}{\cos x} - 1\right)}{x^3} \]
This simplifies to:
\[ \lim_{x \to 0} \frac{\sin x \left(\frac{1 - \cos x}{\cos x}\right)}{x^3} \]

### Step 3: Separate the Limit
We can separate the limit into two parts:
\[ \lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \frac{1 - \cos x}{x^2 \cos x} \]
We know that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), so we focus on the second limit:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2 \cos x} \]

### Step 4: Evaluate the Second Limit
The limit \( \lim_{x \to 0} (1 - \cos x) \) gives \( 0 \) when evaluated directly, leading to the indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{\sin x}{2x} \quad \text{(using L'Hôpital's Rule)} \]

### Step 5: Apply L'Hôpital's Rule Again
Evaluating \( \lim_{x \to 0} \frac{\sin x}{2x} \):
\[ \lim_{x \to 0} \frac{\sin x}{2x} = \frac{1}{2} \]

### Step 6: Combine the Results
Now, substituting back, we have:
\[ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = 1 \cdot \lim_{x \to 0} \frac{1 - \cos x}{x^2 \cos x} = 1 \cdot \frac{1}{2} = \frac{1}{2} \]

### Final Answer
Thus, the limit evaluates to:
\[ \boxed{\frac{1}{2}} \]