Which is `lim_(xto0)(sinx-tanx)/(x)` equal to ?

Options

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

Answer

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

### Step 1: Substitute \( x = 0 \)
First, we will directly substitute \( x = 0 \) into the expression:

\[ \frac{\sin(0) - \tan(0)}{0} = \frac{0 - 0}{0} = \frac{0}{0} \]

This gives us an indeterminate form \( \frac{0}{0} \).

**Hint:** If you encounter \( \frac{0}{0} \) when substituting, consider using L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} \) results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]

Here, let \( f(x) = \sin x - \tan x \) and \( g(x) = x \).

### Step 3: Differentiate the Numerator and Denominator
Now we differentiate the numerator and the denominator:

- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of \( \tan x \) is \( \sec^2 x \).
- The derivative of \( x \) is \( 1 \).

Thus, we have:

\[ f'(x) = \cos x - \sec^2 x \]
\[ g'(x) = 1 \]

### Step 4: Rewrite the Limit
Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{\cos x - \sec^2 x}{1} \]

### Step 5: Substitute \( x = 0 \) Again
Now, we substitute \( x = 0 \) into the new expression:

\[ \cos(0) - \sec^2(0) = 1 - 1 = 0 \]

### Step 6: Apply L'Hôpital's Rule Again
Since we still have \( \frac{0}{0} \), we apply L'Hôpital's Rule again:

Differentiate the numerator again:

- The derivative of \( \cos x \) is \( -\sin x \).
- The derivative of \( \sec^2 x \) is \( 2\sec^2 x \tan x \).

Thus, we have:

\[ f''(x) = -\sin x - 2\sec^2 x \tan x \]

Now, we rewrite the limit:

\[ \lim_{x \to 0} \frac{-\sin x - 2\sec^2 x \tan x}{0} \]

### Step 7: Substitute \( x = 0 \) Again
Now substitute \( x = 0 \):

\[ -\sin(0) - 2\sec^2(0) \tan(0) = 0 - 0 = 0 \]

### Step 8: Final Application of L'Hôpital's Rule
Since we still have \( \frac{0}{0} \), we apply L'Hôpital's Rule one more time.

After differentiating again and substituting \( x = 0 \), we will eventually find that:

\[ \lim_{x \to 0} \frac{\sin x - \tan x}{x} = 0 \]

### Conclusion
Thus, the limit is:

\[ \boxed{0} \]