We need to evaluate the limit:

$\lim_{x \to 0} \frac{1 - \cos x \sqrt{\cos 2x}}{x^2}$

Step 1: Use Trigonometric Approximations

Without derivatives, let's use standard trigonometric identities.

1. Double-Angle Identity for Cosine:
   $\cos 2x = 2\cos^2 x - 1$

   Taking the square root: $\sqrt{\cos 2x} = \sqrt{2\cos^2 x - 1}$

2. Cosine Taylor Expansion (up to $x^2$):
   $\cos x = 1 - \frac{x^2}{2} + O(x^4)$ $\cos 2x = 1 - 2x^2 + O(x^4)$

   So, approximating inside the square root: $\sqrt{\cos 2x} = \sqrt{1 - 2x^2} = 1 - x^2 + O(x^4)$

3. Multiply $\cos x$ and $\sqrt{\cos 2x}$: $\cos x \sqrt{\cos 2x} = \left(1 - \frac{x^2}{2}\right) \cdot \left(1 - x^2\right)$

   Expanding: $= 1 - \frac{x^2}{2} - x^2 + O(x^4)$

   $= 1 - \frac{3x^2}{2} + O(x^4)$

4. Compute $1 - \cos x \sqrt{\cos 2x}$: $1 - \left(1 - \frac{3x^2}{2} + O(x^4)\right) = \frac{3x^2}{2} + O(x^4)$

Step 2: Evaluate the Limit

$\lim_{x \to 0} \frac{1 - \cos x \sqrt{\cos 2x}}{x^2} = \lim_{x \to 0} \frac{\frac{3x^2}{2} + O(x^4)}{x^2}$

$= \lim_{x \to 0} \left(\frac{3}{2} + O(x^2)\right)$

Since $O(x^2)$ vanishes as $x \to 0$, the result is:

$\frac{3}{2}$

Final Answer:

$\frac{3}{2}$

Would you like a deeper breakdown of any step?

Here are five related problems to practice:

1. $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$
2. $\lim_{x \to 0} \frac{\sin x - x}{x^3}$
3. $\lim_{x \to 0} \frac{\tan x - x}{x^3}$
4. $\lim_{x \to 0} \frac{1 - \cos 2x}{x^2}$
5. $\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}$

Tip:

Using small-angle approximations ($\cos x \approx 1 - x^2/2$, $\sin x \approx x$) helps in evaluating limits efficiently without L'Hôpital's Rule.