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.