`lim_(x to 0) (sinx-tanx)/(x)` equal to?

Options

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

Explanation

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

### Step 1: Identify the limit
We start with the limit:
\[ L = \lim_{x \to 0} \frac{\sin x - \tan x}{x} \]

### Step 2: Substitute \( x = 0 \)
Substituting \( x = 0 \) directly into the expression gives:
\[ L = \frac{\sin(0) - \tan(0)}{0} = \frac{0 - 0}{0} = \frac{0}{0} \]
This is an indeterminate form, so we need to apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:
- The derivative of the numerator \( \sin x - \tan x \) is \( \cos x - \sec^2 x \).
- The derivative of the denominator \( x \) is \( 1 \).

Thus, we can rewrite the limit as:
\[ L = \lim_{x \to 0} \frac{\cos x - \sec^2 x}{1} \]

### Step 4: Simplify the expression
We know that \( \sec^2 x = \frac{1}{\cos^2 x} \), so we can rewrite the limit:
\[ L = \lim_{x \to 0} \left( \cos x - \frac{1}{\cos^2 x} \right) \]

### Step 5: Substitute \( x = 0 \) again
Now we substitute \( x = 0 \):
\[ L = \cos(0) - \frac{1}{\cos^2(0)} = 1 - \frac{1}{1^2} = 1 - 1 = 0 \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\sin x - \tan x}{x} = 0 \]