Let's solve each limit one by one.

i. $\lim_{x \to 4} \frac{x^4 - 256}{x - 4}$

First, notice that $x^4 - 256$ can be factored as a difference of squares: $x^4 - 256 = (x^2 - 16)(x^2 + 16)$ And $x^2 - 16$ can further be factored: $x^2 - 16 = (x - 4)(x + 4)$ So: $x^4 - 256 = (x - 4)(x + 4)(x^2 + 16)$ Thus, the limit becomes: $\lim_{x \to 4} \frac{(x - 4)(x + 4)(x^2 + 16)}{x - 4}$ Cancel the $(x - 4)$ terms: $\lim_{x \to 4} (x + 4)(x^2 + 16)$ Now substitute $x = 4$: $(4 + 4)((4)^2 + 16) = 8(16 + 16) = 8 \times 32 = 256$ So, the limit is 256.

ii. $\lim_{x \to 0} \frac{3 - \sqrt{x^2 + 9}}{x^2}$

We can multiply the numerator and denominator by the conjugate of the numerator: $\lim_{x \to 0} \frac{(3 - \sqrt{x^2 + 9})(3 + \sqrt{x^2 + 9})}{x^2(3 + \sqrt{x^2 + 9})}$ This simplifies the numerator: $\lim_{x \to 0} \frac{9 - (x^2 + 9)}{x^2(3 + \sqrt{x^2 + 9})} = \lim_{x \to 0} \frac{-x^2}{x^2(3 + \sqrt{x^2 + 9})}$ Cancel $x^2$ from the numerator and denominator: $\lim_{x \to 0} \frac{-1}{3 + \sqrt{x^2 + 9}}$ Substitute $x = 0$: $\frac{-1}{3 + \sqrt{9}} = \frac{-1}{3 + 3} = \frac{-1}{6}$ So, the limit is $-\frac{1}{6}$.

iii. $\lim_{x \to 1} \frac{x^2 - 2x + 1}{x - 1}$

Notice that $x^2 - 2x + 1$ can be factored as: $x^2 - 2x + 1 = (x - 1)^2$ So the limit becomes: $\lim_{x \to 1} \frac{(x - 1)^2}{x - 1} = \lim_{x \to 1} (x - 1)$ Substitute $x = 1$: $1 - 1 = 0$ Thus, the limit is 0.

iv. $\lim_{t \to \infty} \frac{t^2 - 5t - 9}{2t^4 + 3t^3}$

Divide the numerator and denominator by $t^4$ (the highest power in the denominator): $\lim_{t \to \infty} \frac{\frac{t^2}{t^4} - \frac{5t}{t^4} - \frac{9}{t^4}}{\frac{2t^4}{t^4} + \frac{3t^3}{t^4}} = \lim_{t \to \infty} \frac{\frac{1}{t^2} - \frac{5}{t^3} - \frac{9}{t^4}}{2 + \frac{3}{t}}$ As $t \to \infty$, all terms with $t$ in the denominator go to 0: $\lim_{t \to \infty} \frac{0 - 0 - 0}{2 + 0} = \frac{0}{2} = 0$ So, the limit is 0.

v. $\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}}$

Multiply both the numerator and denominator by the conjugate of the numerator: $\lim_{x \to 0} \frac{(\sqrt{1 + x^2} - \sqrt{1 + x})(\sqrt{1 + x^2} + \sqrt{1 + x})}{(\sqrt{1 + x^3} - \sqrt{1 + x})(\sqrt{1 + x^2} + \sqrt{1 + x})}$ The numerator simplifies to: $(1 + x^2) - (1 + x) = x^2 - x$ So the limit becomes: $\lim_{x \to 0} \frac{x(x - 1)}{(\sqrt{1 + x^3} - \sqrt{1 + x})(\sqrt{1 + x^2} + \sqrt{1 + x})}$ Simplify further, and as $x \to 0$: $\frac{0 \cdot (-1)}{(0)(2)} = 0$ Therefore, the limit is 0.

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Would you like a detailed explanation of any of these solutions or have any questions?

Here are 5 related questions you might explore:

1. How can L'Hôpital's Rule be used to evaluate limits that result in indeterminate forms?
2. What is the geometric interpretation of limits approaching infinity?
3. Can you solve similar limit problems using Taylor series expansion?
4. How do you approach evaluating limits that involve trigonometric functions?
5. What are the common techniques for factoring polynomials to simplify limits?

Tip: Always check if a function can be simplified before applying more advanced limit techniques. Simplifying often avoids unnecessary complexity.