`lim_( x rarr0) (tanx - sinx )/(x^(3))` is equal to

Options

A. `1//2`
B. `-1//2`
C. 0
D. 1

Answer

To solve 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 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 fractions
Next, we 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:
\[ \lim_{x \to 0} \frac{\sin x (1 - \cos x)}{x^3 \cos x} \]

### Step 3: Factor out the limit
We can separate the limit into two parts:
\[ \lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \frac{1 - \cos x}{x^2} \cdot \lim_{x \to 0} \frac{1}{\cos x} \]
We know that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} \frac{1}{\cos x} = 1 \).

### Step 4: Evaluate \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} \)
To evaluate \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} \), we notice that it gives a \( \frac{0}{0} \) form. We can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{\sin x}{2x} = \frac{1}{2} \]

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

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \frac{1}{2} \]

---