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We are tasked with evaluating the following limit:

\[
\lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x}.
\]

Let us break this problem into steps:

### Step 1: Approximation of \(\cos x\)
Near \(x = 0\), we use the Taylor series approximation:
\[
\cos x \approx 1 - \frac{x^2}{2}.
\]

### Step 2: Expand \(\sqrt{\cos x}\) and \(\sqrt[3]{\cos x}\)
Using binomial expansions:
- For \(\sqrt{\cos x} = (\cos x)^{1/2}\):
\[
\sqrt{\cos x} \approx \sqrt{1 - \frac{x^2}{2}} \approx 1 - \frac{x^2}{4}.
\]

- For \(\sqrt[3]{\cos x} = (\cos x)^{1/3}\):
\[
\sqrt[3]{\cos x} \approx \sqrt[3]{1 - \frac{x^2}{2}} \approx 1 - \frac{x^2}{6}.
\]

### Step 3: Simplify the numerator
\[
\sqrt{\cos x} - \sqrt[3]{\cos x} \approx \left(1 - \frac{x^2}{4}\right) - \left(1 - \frac{x^2}{6}\right) = \frac{x^2}{6} - \frac{x^2}{4}.
\]

Simplify further:
\[
\sqrt{\cos x} - \sqrt[3]{\cos x} \approx \frac{x^2}{12}.
\]

### Step 4: Approximation of \(\sin^2 x\)
Near \(x = 0\), \(\sin x \approx x\). Thus:
\[
\sin^2 x \approx x^2.
\]

### Step 5: Simplify the fraction
Substitute into the limit:
\[
\frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} \approx \frac{\frac{x^2}{12}}{x^2}.
\]

Simplify:
\[
\frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} \approx \frac{1}{12}.
\]

### Step 6: Evaluate the limit
As \(x \to 0\), the limit is:
\[
\lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} = \frac{1}{12}.
\]

### Final Answer:
\[
\boxed{\frac{1}{12}}
\]

Would you like a detailed explanation of any specific step? Here are some related questions to expand understanding:

1. How do Taylor series expansions help in solving limits involving trigonometric functions?
2. Can the same technique be applied to other similar problems involving higher-order roots?
3. What is the justification for approximating \(\sin x \approx x\) near zero?
4. How does the binomial theorem aid in expanding functions near their base values?
5. Could L’Hôpital’s Rule be used to evaluate this limit instead?

**Tip:** When solving trigonometric limits, remember that Taylor approximations and simplifications near small angles (\(x \to 0\)) are often the most efficient methods

Evaluate the following limit: \( \lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} \)

Explore how to evaluate the limit \( \lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} \) using Taylor series expansions and approximations. This step-by-step guide is perfect for high school students learning about limits and trigonometric functions.

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