`lim_(x to 0) (1 - cos x)/(x sqrt(x^(2))`

Options

A. `(1)/(2)`
B. `-(1)/(2)`
C. 0
D. Does not exist

Answer

To solve the limit \( \lim_{x \to 0} \frac{1 - \cos x}{x \sqrt{x^2}} \), we will follow these steps:

### Step 1: Rewrite the expression
First, we can simplify the expression in the limit. We know that \( \sqrt{x^2} = |x| \). Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x |x|} \]

### Step 2: Consider the right-hand limit
Next, we will find the right-hand limit as \( x \) approaches 0 from the positive side (\( x \to 0^+ \)). In this case, \( |x| = x \). Therefore, the limit becomes:
\[ \lim_{x \to 0^+} \frac{1 - \cos x}{x^2} \]

### Step 3: Apply the trigonometric limit
Using the known limit \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \), we can evaluate:
\[ \lim_{x \to 0^+} \frac{1 - \cos x}{x^2} = \frac{1}{2} \]

### Step 4: Consider the left-hand limit
Now, we will find the left-hand limit as \( x \) approaches 0 from the negative side (\( x \to 0^- \)). In this case, \( |x| = -x \). Thus, the limit becomes:
\[ \lim_{x \to 0^-} \frac{1 - \cos x}{x (-x)} = \lim_{x \to 0^-} \frac{1 - \cos x}{-x^2} \]

### Step 5: Apply the trigonometric limit again
Using the same known limit:
\[ \lim_{x \to 0^-} \frac{1 - \cos x}{-x^2} = -\frac{1}{2} \]

### Step 6: Compare the limits
Now we have:
- Right-hand limit (RHL): \( \frac{1}{2} \)
- Left-hand limit (LHL): \( -\frac{1}{2} \)

Since RHL \( \neq \) LHL, we conclude that the limit does not exist.

### Final Result
Thus, we can state that:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x \sqrt{x^2}} \text{ does not exist.} \]