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The image you provided contains a mathematical expression:

\[
\lim_{{x \to 0}} \frac{\sin^2(x)}{x^2} > \frac{3x^4 + x^5}{x^6}
\]

Let's analyze and simplify this expression.

### Step 1: Simplify the Left-Hand Side (LHS)
The limit on the left-hand side (LHS) is:

\[
\lim_{{x \to 0}} \frac{\sin^2(x)}{x^2}
\]

Using the small-angle approximation for \(\sin(x)\), where \(\sin(x) \approx x\) when \(x\) is close to 0, we have:

\[
\frac{\sin^2(x)}{x^2} \approx \frac{x^2}{x^2} = 1
\]

Thus, the limit becomes:

\[
\lim_{{x \to 0}} \frac{\sin^2(x)}{x^2} = 1
\]

### Step 2: Simplify the Right-Hand Side (RHS)
The expression on the right-hand side (RHS) is:

\[
\frac{3x^4 + x^5}{x^6}
\]

This can be simplified by dividing both the numerator and denominator by \(x^6\):

\[
\frac{3x^4 + x^5}{x^6} = \frac{3}{x^2} + \frac{1}{x}
\]

As \(x \to 0\), the term \(\frac{3}{x^2}\) and \(\frac{1}{x}\) both tend to infinity. Therefore:

\[
\frac{3x^4 + x^5}{x^6} \to \infty
\]

### Step 3: Compare LHS and RHS
Now, let's compare the LHS and RHS as \(x\) approaches 0:

- LHS: \(1\)
- RHS: \(\infty\)

Since \(1\) is always less than \(\infty\), the inequality:

\[
1 > \infty
\]

is **false**.

### Conclusion
Therefore, the original inequality:

\[
\lim_{{x \to 0}} \frac{\sin^2(x)}{x^2} > \frac{3x^4 + x^5}{x^6}
\]

is not true as \(x \to 0\).

Would you like further details or have any questions?

Here are five related questions to consider:
1. What happens to the inequality if higher-order terms of \(\sin(x)\) are considered?
2. How does the inequality change if the limit is taken as \(x \to \infty\)?
3. Can the small-angle approximation be used for other trigonometric functions in limits?
4. What would the result be if the numerator on the RHS was different?
5. How would you interpret this inequality if the limit was from the left (\(x \to 0^-\)) instead of right?

**Tip:** When working with limits involving trigonometric functions, the small-angle approximation (\(\sin(x) \approx x\) as \(x \to 0\)) is a powerful tool that simplifies the problem.

Find the limit as x approaches 0: \(\lim_{{x \to 0}} \frac{\sin^2(x)}{x^2} > \frac{3x^4 + x^5}{x^6}\)

Understand the detailed analysis of the limit involving trigonometric functions and inequalities as x approaches 0. This solution covers the small-angle approximation, limit properties, and a comparison of results. Ideal for students in Grades 11-12.

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