2. Determine the following limits (if they exist): (2 marks each) \( \begin{array}{ll}\text { a) } \lim _{x \rightarrow 2} \frac{3 x^{2}-4 x}{3-x} & \text { f) } \lim _{y \rightarrow 1} \frac{y+1}{y-1} \\ \text { b) } \lim _{x \rightarrow 4} \frac{x^{2}-x-12}{x-4} & \text { g) } \lim _{h \rightarrow 0} \frac{3 h+h^{2}}{h} \\ \text { c) } \lim _{x \rightarrow 2}\left(3 x+\frac{1}{3 x}\right) & \text { h) } \lim _{h \rightarrow 1} \frac{h^{3}-1}{h-1} \\ \text { d) } \lim _{x \rightarrow 0} \frac{1}{x} & \text { i) } \lim _{x \rightarrow 3} \frac{\sqrt{x}-\sqrt{3}}{x-3} \\ \text { e) } \lim _{y \rightarrow 1} \frac{y-1}{y+1} & \end{array} \)

Let's evaluate each limit one by one. ### a) \( \lim _{x \rightarrow 2} \frac{3 x^{2}-4 x}{3-x} \) First, substitute \( x = 2 \): \[ \frac{3(2)^2 - 4(2)}{3 - 2} = \frac{3(4) - 8}{1} = \frac{12 - 8}{1} = \frac{4}{1} = 4 \] **Limit: 4** ### b) \( \lim _{x \rightarrow 4} \frac{x^{2}-x-12}{x-4} \) Factor the numerator: \[ x^2 - x - 12 = (x - 4)(x + 3) \] Thus, \[ \frac{(x - 4)(x + 3)}{x - 4} = x + 3 \quad \text{for } x \neq 4 \] Now substitute \( x = 4 \): \[ 4 + 3 = 7 \] **Limit: 7** ### c) \( \lim _{x \rightarrow 2}\left(3 x+\frac{1}{3 x}\right) \) Substituting \( x = 2 \): \[ 3(2) + \frac{1}{3(2)} = 6 + \frac{1}{6} = 6 + 0.1667 = 6.1667 \approx \frac{37}{6} \] **Limit: \( \frac{37}{6} \)** ### d) \( \lim _{x \rightarrow 0} \frac{1}{x} \) As \( x \) approaches 0, \( \frac{1}{x} \) approaches \( +\infty \) from the right and \( -\infty \) from the left. Therefore, the limit does not exist. **Limit: Does not exist** ### e) \( \lim _{y \rightarrow 1} \frac{y-1}{y+1} \) Substituting \( y = 1 \): \[ \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 \] **Limit: 0** ### f) \( \lim _{y \rightarrow 1} \frac{y+1}{y-1} \) Substituting \( y = 1 \): \[ \frac{1 + 1}{1 - 1} = \frac{2}{0} \] This approaches \( +\infty \) from the right and \( -\infty \) from the left. Therefore, the limit does not exist. **Limit: Does not exist** ### g) \( \lim _{h \rightarrow 0} \frac{3 h+h^{2}}{h} \) Factor out \( h \): \[ \frac{h(3 + h)}{h} = 3 + h \quad \text{for } h \neq 0 \] Now substitute \( h = 0 \): \[ 3 + 0 = 3 \] **Limit: 3** ### h) \( \lim _{h \rightarrow 1} \frac{h^{3}-1}{h-1} \) Factor the numerator: \[ h^3 - 1 = (h - 1)(h^2 + h + 1) \] Thus, \[ \frac{(h - 1)(h^2 + h + 1)}{h - 1} = h^2 + h + 1 \quad \text{for } h \neq 1 \] Now substitute \( h = 1 \): \[ 1^2 + 1 + 1 = 3 \] **Limit: 3** ### i) \( \lim _{x \rightarrow 3} \frac{\sqrt{x}-\sqrt{3}}{x-3} \) This is an indeterminate form \( \frac{0}{0} \). We can use L'Hôpital's Rule: \[ \text{Differentiate the numerator and denominator:} \] \[ \text{Numerator: } \frac{1}{2\sqrt{x}}, \quad \text{Denominator: } 1 \] Now apply L'Hôpital's Rule: \[ \lim_{x \to 3} \frac{\frac{1}{2\sqrt{x}}}{1} = \frac{1}{2\sqrt{3}} \] **Limit: \( \