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

Answer

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

### Step 1: Substitute \( x = 0 \)
First, we substitute \( x = 0 \) into the expression to check if it results in an indeterminate form.

\[ \cos(2 \cdot 0) - 1 = \cos(0) - 1 = 1 - 1 = 0 \]
\[ \cos(0) - 1 = 1 - 1 = 0 \]

Thus, we have \( \frac{0}{0} \), which is an indeterminate form.

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

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator.

**Numerator:**
\[ \frac{d}{dx}(\cos 2x - 1) = -\sin 2x \cdot 2 = -2\sin 2x \]

**Denominator:**
\[ \frac{d}{dx}(\cos x - 1) = -\sin x \]

Now, we rewrite the limit using these derivatives:
\[ \lim_{x \to 0} \frac{-2 \sin 2x}{-\sin x} = \lim_{x \to 0} \frac{2 \sin 2x}{\sin x} \]

**Hint:** Remember to differentiate both the numerator and denominator separately.

### Step 3: Simplify the Expression
Using the identity \( \sin 2x = 2 \sin x \cos x \), we can rewrite the limit:
\[ \lim_{x \to 0} \frac{2(2 \sin x \cos x)}{\sin x} = \lim_{x \to 0} \frac{4 \sin x \cos x}{\sin x} \]

Now, we can cancel \( \sin x \) from the numerator and denominator (as long as \( x \neq 0 \)):
\[ \lim_{x \to 0} 4 \cos x \]

**Hint:** Use trigonometric identities to simplify the expression.

### Step 4: Evaluate the Limit
Now we can substitute \( x = 0 \) into the simplified limit:
\[ 4 \cos(0) = 4 \cdot 1 = 4 \]

Thus, the limit is:
\[ \lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1} = 4 \]

**Final Answer:**
\[ \boxed{4} \]