1. **Problem:** Evaluate the limits and prove logical equivalence: (a) Evaluate $$\lim_{x \to 0} x^2 \sin \frac{1}{x}$$, $$\lim_{x \to 0} \sin \frac{1}{x}$$, and $$\lim_{x \to 0} x$$. (b) Prove for any two statements p and q, $$\sim(p \lor q) \equiv \sim p \land \sim q$$. (c) Define a statement and explain how it differs from a sentence in logic. (d) Explain the relationship between the existence of limits and continuity at a point. (e) For the piecewise function $$f(x) = \begin{cases} \tan x - \cot x, & x \neq \frac{\pi}{4} \\ a, & x = \frac{\pi}{4} \end{cases}$$ find a so that f is continuous at $$x=\frac{\pi}{4}$$. (f) Evaluate $$\lim_{x \to y} \frac{\tan x - \tan y}{x - y}$$. (g) Find by first principles the derivative of $$g(x) = \cot(b - ax)$$. 2. **Step-by-step solution:** 1. Limits: - $$\lim_{x \to 0} x^2 \sin \frac{1}{x}$$: Since $$|\sin \frac{1}{x}| \leq 1$$, we have $$|x^2 \sin \frac{1}{x}| \leq |x^2|$$. As $$x \to 0$$, $$x^2 \to 0$$, so by squeeze theorem, $$\lim_{x \to 0} x^2 \sin \frac{1}{x} = 0.$$ - $$\lim_{x \to 0} \sin \frac{1}{x}$$: The argument $$\frac{1}{x}$$ oscillates wildly near 0, so the limit does not exist. - $$\lim_{x \to 0} x = 0.$$ 2. Proof of logical equivalence: To prove $$\sim(p \lor q) \equiv \sim p \land \sim q$$, use truth tables or logical laws: - Using De Morgan's laws: $$\sim(p \lor q) = \sim p \land \sim q.$$ This shows both sides are equivalent. 3. Statement vs Sentence: - A **statement** in logic is a sentence that is either true or false. - A **sentence** may be meaningful but not necessarily have a truth value. Thus, every statement is a sentence, but not every sentence is a statement. 4. Relationship between limits and continuity: - A function $$f$$ is **continuous at** $$x=c$$ if $$\lim_{x \to c} f(x) = f(c).$$ - The limit must exist, and the function's value at $$c$$ must equal that limit. 5. Continuity of $$f(x) = \tan x - \cot x$$ at $$x=\frac{\pi}{4}$$: - Compute $$\lim_{x \to \frac{\pi}{4}} f(x)$$: $$\tan \frac{\pi}{4} = 1, \quad \cot \frac{\pi}{4} = 1,$$ so the numerator approaches $$1-1=0$$. - But examine closely: $$f(x) = \tan x - \cot x = \frac{\sin x}{\cos x} - \frac{\cos x}{\sin x} = \frac{\sin^2 x - \cos^2 x}{\sin x \cos x}.$$ At $$x=\frac{\pi}{4}$$: $$\sin^2 \frac{\pi}{4} = \cos^2 \frac{\pi}{4} = \frac{1}{2},$$ thus numerator $$= \frac{1}{2} - \frac{1}{2} = 0,$$ and denominator $$\sin \frac{\pi}{4} \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} = \frac{1}{2}.$$ Hence, $$\lim_{x \to \frac{\pi}{4}} f(x) = \frac{0}{\frac{1}{2}} = 0.$$ For continuity, set $$a = 0$$. 6. Evaluate $$\lim_{x \to y} \frac{\tan x - \tan y}{x - y}$$: By definition of derivative, this is $$\frac{d}{dx} \tan x \bigg|_{x=y} = \sec^2 y.$$ 7. Derivative by first principles of $$g(x) = \cot(b - a x)$$: - By definition: $$g'(x) = \lim_{h \to 0} \frac{\cot(b - a (x + h)) - \cot(b - a x)}{h}.$$ - Using derivative of cotangent: $$\frac{d}{dx} \cot u = -\csc^2 u \cdot \frac{du}{dx}.$$ - Here, $$u = b - ax$$, so $$\frac{du}{dx} = -a.$$ Therefore, $$g'(x) = -\csc^2(b - ax) \times (-a) = a \csc^2(b - ax).$$ **Final answers:** $$\lim_{x \to 0} x^2 \sin \frac{1}{x} = 0,$$ $$\lim_{x \to 0} \sin \frac{1}{x} \text{ does not exist},$$ $$\lim_{x \to 0} x = 0,$$ $$\sim(p \lor q) \equiv \sim p \land \sim q,$$ $$a = 0$$ for continuity, $$\lim_{x \to y} \frac{\tan x - \tan y}{x - y} = \sec^2 y,$$ and $$\frac{d}{dx} \cot(b - a x) = a \csc^2(b - a x).$$