. Find the Limits of the following functions: i. 𝑦 = 𝑓(𝑥) = 𝑥 2sin( 1 𝑥2 ) 𝑐𝑜𝑠𝑥 as 𝑥 → 0. ii. lim𝑥→0 8𝑥 3−27 4𝑥 2+6𝑥+9 iii. lim𝑥→2 √1+√2+𝑥−√3 𝑥−2 iv. lim𝑥→0 𝑡𝑎𝑛𝑥 𝑠𝑖𝑛𝑥 v. lim𝑥→0 |𝑥|, where |𝑥| = { 𝑥 𝑖𝑓 𝑥 ≥ 0 −𝑥 𝑖𝑓 𝑥 < 0

Let's analyze and compute the limits for each function step by step.

### i. \(\lim_{x \to 0} \, f(x) = x^2 \sin\left(\frac{1}{x^2}\right) \cos(x)\)

- **Approach**: As \( x \to 0 \), we analyze the behavior of each factor:
  - \( \sin\left(\frac{1}{x^2}\right) \) oscillates between \(-1\) and \(1\) as \(x \to 0\).
  - \( \cos(x) \) tends to \(1\) as \( x \to 0 \) because \(\cos(0) = 1\).
  - The key is that \( x^2 \to 0 \) as \( x \to 0 \), which will dominate the behavior.
  
  Since \( \sin\left(\frac{1}{x^2}\right) \) is bounded by \([-1, 1]\), we use the **Squeeze Theorem**:
  \[
  -x^2 \leq x^2 \sin\left(\frac{1}{x^2}\right) \cos(x) \leq x^2
  \]
  As \( x^2 \to 0 \), both bounds go to 0, implying:
  \[
  \lim_{x \to 0} x^2 \sin\left(\frac{1}{x^2}\right) \cos(x) = 0
  \]

### ii. \(\lim_{x \to 0} \frac{8x^3 - 27}{4x^2 + 6x + 9}\)

- **Approach**: This is a rational function. Let's first evaluate the expression directly at \(x = 0\):
  \[
  \text{At } x = 0: \frac{8(0)^3 - 27}{4(0)^2 + 6(0) + 9} = \frac{-27}{9} = -3
  \]
  Hence, the limit is:
  \[
  \lim_{x \to 0} \frac{8x^3 - 27}{4x^2 + 6x + 9} = -3
  \]

### iii. \(\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2}\)

- **Approach**: This is of the indeterminate form \(\frac{0}{0}\), so we apply **L'Hopital's Rule**.
  - First, differentiate the numerator and the denominator separately:
    - Numerator: \( f(x) = \sqrt{1 + \sqrt{2 + x}} \). The derivative using the chain rule is:
      \[
      f'(x) = \frac{1}{2\sqrt{1 + \sqrt{2 + x}}} \cdot \frac{1}{2\sqrt{2 + x}}
      \]
    - Denominator: \( g(x) = x - 2 \). The derivative is simply:
      \[
      g'(x) = 1
      \]
  - Now, applying L'Hopital's Rule:
    \[
    \lim_{x \to 2} \frac{\frac{1}{2\sqrt{1 + \sqrt{2 + x}}} \cdot \frac{1}{2\sqrt{2 + x}}}{1}
    \]
    Substituting \(x = 2\):
    \[
    \frac{1}{2\sqrt{1 + \sqrt{2 + 2}}} \cdot \frac{1}{2\sqrt{2 + 2}} = \frac{1}{2\sqrt{1 + 2}} \cdot \frac{1}{2\sqrt{4}} = \frac{1}{2\sqrt{3}} \cdot \frac{1}{4} = \frac{1}{8\sqrt{3}}
    \]
  Hence, the limit is \( \frac{1}{8\sqrt{3}} \).

### iv. \(\lim_{x \to 0} \frac{\tan(x)}{\sin(x)}\)

- **Approach**: We use the small-angle approximations for \( \tan(x) \approx x \) and \( \sin(x) \approx x \) as \( x \to 0 \):
  \[
  \lim_{x \to 0} \frac{\tan(x)}{\sin(x)} = \lim_{x \to 0} \frac{x}{x} = 1
  \]
  Hence, the limit is:
  \[
  1
  \]

### v. \(\lim_{x \to 0} |x|\)

- **Approach**: The absolute value function \( |x| \) is continuous and equal to \( x \) if \( x \geq 0 \) and \(-x\) if \( x < 0 \). In either case, as \( x \to 0 \), the function approaches 0.
  \[
  \lim_{x \to 0} |x| = 0
  \]

### Summary of Limits:
1. \( \lim_{x \to 0} x^2 \sin\left(\frac{1}{x^2}\right) \cos(x) = 0 \)
2. \( \lim_{x \to 0} \frac{8x^3 - 27}{4x^2 + 6x + 9} = -3 \)
3. \( \lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2} = \frac{1}{8\sqrt{3}} \)
4. \( \lim_{x \to 0} \frac{\tan(x)}{\sin(x)} = 1 \)
5. \( \lim_{x \to 0} |x| = 0 \)

Would you like further explanations or details on any of these limits?

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Here are 5 related questions for practice:
1. What is the limit of \( \frac{\sin(x)}{x} \) as \( x \to 0 \)?
2. Evaluate \( \lim_{x \to \infty} \frac{1}{x^2} \).
3. Compute \( \lim_{x \to 0} \frac{\cos(x) - 1}{x^2} \).
4. What is \( \lim_{x \to \infty} \frac{\log(x)}{x} \)?
5. How does L'Hopital's Rule help in evaluating indeterminate limits?

**Tip**: When dealing with limits involving oscillating functions, the Squeeze Theorem is often a powerful tool to determine the limit.

Find the Limits of the following functions: i. 𝑦 = 𝑓(𝑥) = 𝑥^2 sin(1/𝑥^2) cos(𝑥) as 𝑥 → 0. ii. lim𝑥→0 (8𝑥^3 − 27)/(4𝑥^2 + 6𝑥 + 9) iii. lim𝑥→2 (√(1+√(2+𝑥)) − √3)/(𝑥−2) iv. lim𝑥→0 tan(𝑥)/sin(𝑥) v. lim𝑥→0 |𝑥|, where |𝑥| = { 𝑥 if 𝑥 ≥ 0, −𝑥 if 𝑥 < 0.

Solve various limit problems including trigonometric, rational, and square root functions using advanced techniques like L'Hopital's Rule and the Squeeze Theorem. This detailed guide covers key mathematical concepts and step-by-step solutions suitable for high school students in Grades 11-12.

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