Thanks for sharing the image! I'll solve the provided limits step by step. Let's work through each problem:

3(a)

$\lim_{x \to 2} \frac{x^5 - 32}{x - 2}$
Solution:
The numerator $x^5 - 32$ can be factored using the difference of powers formula:
$x^5 - 32 = (x - 2)(x^4 + 2x^3 + 4x^2 + 8x + 16).$
Substituting this into the fraction:
$\frac{x^5 - 32}{x - 2} = x^4 + 2x^3 + 4x^2 + 8x + 16 \quad \text{for } x \neq 2.$
Now, directly substitute $x = 2$:
$\lim_{x \to 2} \frac{x^5 - 32}{x - 2} = 2^4 + 2(2^3) + 4(2^2) + 8(2) + 16 = 16 + 16 + 16 + 16 + 16 = 80.$

Answer: $80$

3(b)

$\lim_{x \to a} \frac{\frac{1}{x^3} - \frac{1}{a^3}}{\frac{1}{x} - \frac{1}{a}}$
Solution:
Simplify the fractions:
$\frac{\frac{1}{x^3} - \frac{1}{a^3}}{\frac{1}{x} - \frac{1}{a}} = \frac{\frac{a^3 - x^3}{x^3 a^3}}{\frac{a - x}{ax}}.$
Simplify further:
$= \frac{a^3 - x^3}{x^3 a^3} \cdot \frac{ax}{a - x}.$
Notice $a^3 - x^3 = (a - x)(a^2 + ax + x^2)$. Substituting this:
$= \frac{(a - x)(a^2 + ax + x^2)}{x^3 a^3} \cdot \frac{ax}{a - x}.$
Cancel out $a - x$:
$= \frac{a^2 + ax + x^2}{x^3 a^3} \cdot ax.$
Substitute $x = a$:
$= \frac{a^2 + a^2 + a^2}{a^3 a^3} \cdot a = \frac{3a^2}{a^6} \cdot a = \frac{3}{a^3}.$

Answer: $\frac{3}{a^3}$

3(c)

[ \lim_{x \to 64} \frac{\sqrt[6]{x} - 2}{\sqrt[3]{x} - 4} ]
Solution:
Let $x = t^6$, so $\sqrt[6]{x} = t$ and $\sqrt[3]{x} = t^2$. Substituting these:
[ \frac{\sqrt[6]{x} - 2}{\sqrt[3]{x} - 4} = \frac{t - 2}{t^2 - 4}. ]
Factor $t^2 - 4 = (t - 2)(t + 2)$:
$= \frac{1}{t + 2} \quad \text{for } t \neq 2.$
Substitute $t = 2$:
[ \lim_{x \to 64} \frac{\sqrt[6]{x} - 2}{\sqrt[3]{x} - 4} = \frac{1}{2 + 2} = \frac{1}{4}. ]

Answer: $\frac{1}{4}$

3(d)

$\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}$
Solution:
Factor both numerator and denominator:
$x^3 - 64 = (x - 4)(x^2 + 4x + 16), \quad x^2 - 16 = (x - 4)(x + 4).$
Simplify:
$\frac{x^3 - 64}{x^2 - 16} = \frac{x^2 + 4x + 16}{x + 4} \quad \text{for } x \neq 4.$
Substitute $x = 4$:
$\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16} = \frac{4^2 + 4(4) + 16}{4 + 4} = \frac{16 + 16 + 16}{8} = \frac{48}{8} = 6.$

Answer: $6$

3(e)

$\lim_{x \to b} \frac{x^{5/2} - b^{5/2}}{x - b}$
Solution:
This is a derivative in disguise! Using the definition of the derivative:
$\lim_{x \to b} \frac{x^{5/2} - b^{5/2}}{x - b} = \frac{d}{dx} \left( x^{5/2} \right) \Big|_{x = b}.$
Differentiate $x^{5/2}$:
$\frac{d}{dx} x^{5/2} = \frac{5}{2}x^{3/2}.$
Substitute $x = b$:
$\lim_{x \to b} \frac{x^{5/2} - b^{5/2}}{x - b} = \frac{5}{2}b^{3/2}.$

Answer: $\frac{5}{2}b^{3/2}$

Let me know if you need further explanations for any part!

Related Questions:

1. What happens if the denominator approaches zero but the numerator doesn't?
2. How can L'Hôpital's Rule be applied to limits involving indeterminate forms?
3. Why does factoring simplify many limit problems?
4. How can limits define the derivative of a function?
5. What are other common forms where direct substitution fails in limits?

Tip: Always try direct substitution first; if it results in an indeterminate form, explore simplifications or apply advanced methods like factoring or L'Hôpital's Rule.