This exercise involves limits and identifying points of discontinuity for a piecewise function. I'll break it down into two parts.

Part 1: Finding the limits

The first set of problems asks to compute limits without using L'Hopital's Rule.

a) $\lim_{x \to \infty} \frac{3x^3 + 4x}{8x^3 - 4x^2 + 5}$

• As $x \to \infty$, the terms with the highest degree in $x$ will dominate both the numerator and denominator.
• The leading term in the numerator is $3x^3$ and in the denominator, it's $8x^3$.
• So, $\lim_{x \to \infty} \frac{3x^3 + 4x}{8x^3 - 4x^2 + 5} = \frac{3}{8}$.

b) $\lim_{x \to 1} \frac{x^2 - 4x + 3}{x^2 - 1}$

• First, factor the numerator and denominator: $\text{Numerator: } x^2 - 4x + 3 = (x - 1)(x - 3)$ $\text{Denominator: } x^2 - 1 = (x - 1)(x + 1)$
• The factor $(x - 1)$ cancels out: $\lim_{x \to 1} \frac{(x - 1)(x - 3)}{(x - 1)(x + 1)} = \lim_{x \to 1} \frac{x - 3}{x + 1}$
• Now substitute $x = 1$: $\frac{1 - 3}{1 + 1} = \frac{-2}{2} = -1.$

c) $\lim_{x \to \infty} \sqrt{x - \sqrt{x^2 + 7x}}$

• To simplify, multiply by the conjugate: $\sqrt{x - \sqrt{x^2 + 7x}} \cdot \frac{\sqrt{x + \sqrt{x^2 + 7x}}}{\sqrt{x + \sqrt{x^2 + 7x}}}$
• After simplification, the limit tends to $-\frac{7}{2}$ as $x \to \infty$.

d) $\lim_{x \to 0} \frac{\sqrt{1 - \cos(5x)}}{2x}$

• Use a small angle approximation: $\cos(5x) \approx 1 - \frac{(5x)^2}{2}$ as $x \to 0$.
• The expression becomes: $\frac{\sqrt{1 - \left(1 - \frac{25x^2}{2}\right)}}{2x} = \frac{\sqrt{\frac{25x^2}{2}}}{2x} = \frac{\frac{5x}{\sqrt{2}}}{2x} = \frac{5}{2\sqrt{2}}.$

e) $\lim_{x \to \infty} \left( \frac{2x - 1}{2x + 2} \right)^{3x}$

• Simplify the fraction: $\frac{2x - 1}{2x + 2} = 1 - \frac{3}{2x + 2}$
• This approaches 1 as $x \to \infty$, so the overall expression approaches $1^{\infty}$, which tends to $e^{-3}$.

Part 2: Identifying points of discontinuity for the piecewise function

The function is given as: