1. Evaluate the limits using limit laws: (i) Find $$\lim_{x \to 2} \frac{2x^3 - 3x + 1}{x^3 + 4}$$ Step 1: Substitute $x=2$ directly since polynomials are continuous: $$\frac{2(2)^3 - 3(2) + 1}{(2)^3 + 4} = \frac{2(8) - 6 + 1}{8 + 4} = \frac{16 - 6 + 1}{12} = \frac{11}{12}$$ Answer: $$\frac{11}{12}$$ (ii) Find $$\lim_{t \to 0} \frac{\sqrt{1+t} - \sqrt{1-t}}{t}$$ Step 1: Multiply numerator and denominator by the conjugate to simplify: $$\frac{\sqrt{1+t} - \sqrt{1-t}}{t} \times \frac{\sqrt{1+t} + \sqrt{1-t}}{\sqrt{1+t} + \sqrt{1-t}} = \frac{(1+t) - (1-t)}{t(\sqrt{1+t} + \sqrt{1-t})} = \frac{2t}{t(\sqrt{1+t} + \sqrt{1-t})}$$ Step 2: Cancel $t$: $$\frac{\cancel{2t}}{\cancel{t}(\sqrt{1+t} + \sqrt{1-t})} = \frac{2}{\sqrt{1+t} + \sqrt{1-t}}$$ Step 3: Substitute $t=0$: $$\frac{2}{\sqrt{1+0} + \sqrt{1-0}} = \frac{2}{1 + 1} = 1$$ Answer: $$1$$ (iii) Find $$\lim_{x \to 3} \frac{\frac{1}{x+3} + 1}{x+3}$$ Step 1: Simplify numerator: $$\frac{1}{x+3} + 1 = \frac{1 + (x+3)}{x+3} = \frac{x+4}{x+3}$$ Step 2: Substitute into limit expression: $$\frac{\frac{x+4}{x+3}}{x+3} = \frac{x+4}{(x+3)^2}$$ Step 3: Substitute $x=3$: $$\frac{3+4}{(3+3)^2} = \frac{7}{6^2} = \frac{7}{36}$$ Answer: $$\frac{7}{36}$$ (iv) Find $$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\sin \theta}$$ Step 1: Use trigonometric identity $1 - \cos \theta = 2 \sin^2(\theta/2)$: $$\frac{1 - \cos \theta}{\sin \theta} = \frac{2 \sin^2(\theta/2)}{\sin \theta}$$ Step 2: Rewrite denominator using double angle formula $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$: $$\frac{2 \sin^2(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)} = \frac{\cancel{2} \sin^2(\theta/2)}{\cancel{2} \sin(\theta/2) \cos(\theta/2)} = \frac{\sin(\theta/2)}{\cos(\theta/2)} = \tan(\theta/2)$$ Step 3: Take limit as $\theta \to 0$: $$\lim_{\theta \to 0} \tan(\theta/2) = 0$$ Answer: $$0$$ 2. Show $$\lim_{x \to a} \frac{p(x)}{q(x)} = \frac{p(a)}{q(a)}$$ for polynomials $p(x), q(x)$ and real $a$ with $q(a) \neq 0$. Step 1: Polynomials are continuous functions, so: $$\lim_{x \to a} p(x) = p(a)$$ and $$\lim_{x \to a} q(x) = q(a)$$ Step 2: Since $q(a) \neq 0$, limit of quotient is quotient of limits: $$\lim_{x \to a} \frac{p(x)}{q(x)} = \frac{\lim_{x \to a} p(x)}{\lim_{x \to a} q(x)} = \frac{p(a)}{q(a)}$$ 3. Use squeeze theorem: (i) Given $$4x - 9 \leq f(x) \leq x^2 - 4x + 7$$, find $$\lim_{x \to 4} f(x)$$ Step 1: Find limits of bounding functions at $x=4$: $$\lim_{x \to 4} (4x - 9) = 4(4) - 9 = 16 - 9 = 7$$ $$\lim_{x \to 4} (x^2 - 4x + 7) = 16 - 16 + 7 = 7$$ Step 2: Since both limits equal 7 and $f(x)$ is squeezed between them, by squeeze theorem: $$\lim_{x \to 4} f(x) = 7$$ (ii) Find $$\lim_{x \to 0} \sqrt{x^2 - x^2 + 1} \sin(\pi/x)$$ Step 1: Simplify inside the square root: $$\sqrt{x^2 - x^2 + 1} = \sqrt{1} = 1$$ Step 2: So limit becomes: $$\lim_{x \to 0} 1 \cdot \sin(\pi/x)$$ Step 3: Since $\sin(\pi/x)$ oscillates between -1 and 1, but multiplied by 1, the function oscillates between -1 and 1. Step 4: Use squeeze theorem with bounds: $$-1 \leq \sin(\pi/x) \leq 1$$ Step 5: Multiply by 1: $$-1 \leq 1 \cdot \sin(\pi/x) \leq 1$$ Step 6: Limits of bounds as $x \to 0$ are both 1 and -1, so limit does not exist because the function oscillates. Answer: Limit does not exist. 4. Use $\varepsilon$-$\delta$ definition: (i) Prove $$\lim_{x \to -4} -4 \left( \frac{1}{2} x - 1 \right) = -3$$ Step 1: Define function: $$f(x) = -4 \left( \frac{1}{2} x - 1 \right) = -2x + 4$$ Step 2: Compute $f(-4)$: $$f(-4) = -2(-4) + 4 = 8 + 4 = 12$$ Step 3: The problem states limit is -3, but direct substitution gives 12, so likely a typo or misinterpretation. Assuming the problem meant $$\lim_{x \to -4} -4 \left( \frac{1}{2} x - 1 \right) = 12$$. Step 4: For any $\varepsilon > 0$, choose $\delta = \frac{\varepsilon}{2}$. Step 5: If $|x + 4| < \delta$, then: $$|f(x) - 12| = |-2x + 4 - 12| = |-2x - 8| = 2|x + 4| < 2 \delta = \varepsilon$$ Hence, limit is 12. (ii) Prove $$\lim_{x \to 0} \sqrt{x^2} = 0$$ Step 1: Note that $$\sqrt{x^2} = |x|$$. Step 2: For any $\varepsilon > 0$, choose $\delta = \varepsilon$. Step 3: If $|x - 0| < \delta$, then: $$|\sqrt{x^2} - 0| = |x| < \delta = \varepsilon$$ Hence, limit is 0. Final answers: (i) $$\frac{11}{12}$$ (ii) $$1$$ (iii) $$\frac{7}{36}$$ (iv) $$0$$ 2. $$\lim_{x \to a} \frac{p(x)}{q(x)} = \frac{p(a)}{q(a)}$$ 3.(i) $$7$$ 3.(ii) Limit does not exist 4.(i) $$12$$ (assuming correction) 4.(ii) $$0$$