`lim_(xrarr0)(tanx-sinx)/(x^3)=`

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. `(1)/(8)`
D. none of these

Explanation

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

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

### Step 2: Combine the terms in the numerator
Next, we can combine the terms in the numerator:
\[ \frac{\sin x}{\cos x} - \sin x = \sin x \left( \frac{1}{\cos x} - 1 \right) = \sin x \left( \frac{1 - \cos x}{\cos x} \right) \]
Thus, the limit becomes:
\[ \lim_{x \to 0} \frac{\sin x \cdot (1 - \cos x)}{x^3 \cos x} \]

### Step 3: Factor out \( \sin x \)
We can factor out \( \sin x \) from the limit:
\[ \lim_{x \to 0} \left( \frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2} \cdot \frac{1}{\cos x} \right) \]

### Step 4: Evaluate the limits separately
Now we can evaluate the limits separately:
1. \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)
2. For \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} \), we can use the Taylor series expansion of \( \cos x \):
\[ 1 - \cos x \approx \frac{x^2}{2} \text{ as } x \to 0 \]
Therefore,
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{2}}{x^2} = \frac{1}{2} \]
3. Lastly, \( \lim_{x \to 0} \frac{1}{\cos x} = 1 \).

### Step 5: Combine the results
Now we can combine the results:
\[ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = 1 \cdot \frac{1}{2} \cdot 1 = \frac{1}{2} \]

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