1. **Evaluate** \(\lim_{x \to 0} \frac{a^{x} - b^{x}}{x}\). Step 1: Recall the exponential limit property: \(\lim_{x \to 0} \frac{c^{x} - 1}{x} = \ln c\). Step 2: Rewrite the expression: $$\lim_{x \to 0} \frac{a^{x} - b^{x}}{x} = \lim_{x \to 0} \frac{a^{x} - 1 + 1 - b^{x}}{x} = \lim_{x \to 0} \left(\frac{a^{x} - 1}{x} - \frac{b^{x} - 1}{x}\right)$$ Step 3: Apply the limit property: $$= \ln a - \ln b = \ln \left(\frac{a}{b}\right)$$ **Answer:** \(\lim_{x \to 0} \frac{a^{x} - b^{x}}{x} = \ln \left(\frac{a}{b}\right)\). 2. **Calculate:** (i) \(\lim_{x \to \infty} \frac{2x^{5} - 3x^{2} + 5}{3x^{5} + 2x^{2} + 8}\) Step 1: Divide numerator and denominator by \(x^{5}\): $$\lim_{x \to \infty} \frac{2 - 3x^{-3} + 5x^{-5}}{3 + 2x^{-3} + 8x^{-5}}$$ Step 2: As \(x \to \infty\), terms with negative powers go to 0: $$= \frac{2}{3}$$ (ii) \(\lim_{x \to \infty} \frac{3x^{2} - 4x + 1}{x^{2} - 4x + 3}\) Step 1: Divide numerator and denominator by \(x^{2}\): $$\lim_{x \to \infty} \frac{3 - 4x^{-1} + x^{-2}}{1 - 4x^{-1} + 3x^{-2}}$$ Step 2: As \(x \to \infty\), terms with negative powers go to 0: $$= \frac{3}{1} = 3$$ 3. **Show that** \(\lim_{x \to 2} \frac{|x - 2|}{x - 2}\) **does not exist.** Step 1: Consider left-hand limit (\(x \to 2^{-}\)): $$\frac{|x - 2|}{x - 2} = \frac{-(x - 2)}{x - 2} = -1$$ Step 2: Consider right-hand limit (\(x \to 2^{+}\)): $$\frac{|x - 2|}{x - 2} = \frac{x - 2}{x - 2} = 1$$ Step 3: Since left and right limits are not equal, the limit does not exist. 4. **Evaluate** \(\lim_{x \to 2} \frac{\sqrt{3 - x} - 1}{2 - x}\). Step 1: Substitute \(x = 2\) directly: $$\frac{\sqrt{3 - 2} - 1}{2 - 2} = \frac{1 - 1}{0} = \frac{0}{0}$$ (indeterminate form) Step 2: Rationalize numerator: $$\frac{\sqrt{3 - x} - 1}{2 - x} \times \frac{\sqrt{3 - x} + 1}{\sqrt{3 - x} + 1} = \frac{3 - x - 1}{(2 - x)(\sqrt{3 - x} + 1)} = \frac{2 - x}{(2 - x)(\sqrt{3 - x} + 1)}$$ Step 3: Cancel \(2 - x\): $$= \frac{1}{\sqrt{3 - x} + 1}$$ Step 4: Substitute \(x = 2\): $$= \frac{1}{\sqrt{1} + 1} = \frac{1}{2}$$ 5. **Examine continuity of** \(f(x) = x^{3}\) **at** \(x = 2\). Step 1: Calculate \(f(2) = 2^{3} = 8\). Step 2: Calculate \(\lim_{x \to 2} f(x) = \lim_{x \to 2} x^{3} = 8\). Step 3: Since \(f(2) = \lim_{x \to 2} f(x)\), \(f\) is continuous at \(x = 2\). 6. **Discuss continuity of** \(f(x) = \begin{cases} 2x - 1, & x < 0 \\ 2x + 1, & x \geq 0 \end{cases}\) **at** \(x = 0\). Step 1: Calculate left limit: $$\lim_{x \to 0^{-}} f(x) = 2(0) - 1 = -1$$ Step 2: Calculate right limit: $$\lim_{x \to 0^{+}} f(x) = 2(0) + 1 = 1$$ Step 3: Since left and right limits are not equal, \(f\) is not continuous at \(x = 0\). 7. **Find** \(k\) **for continuity of** \(f(x) = \begin{cases} kx + 5, & x \leq 2 \\ x - 1, & x > 2 \end{cases}\) **at** \(x = 2\). Step 1: Calculate left limit and value: $$f(2) = k(2) + 5 = 2k + 5$$ Step 2: Calculate right limit: $$\lim_{x \to 2^{+}} f(x) = 2 - 1 = 1$$ Step 3: For continuity, set equal: $$2k + 5 = 1 \implies 2k = -4 \implies k = -2$$ 8. **Show differentiability of** \(f(x) = \begin{cases} x^{2} \sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}\) **at** \(x = 0\). Step 1: Check continuity at 0: $$\lim_{x \to 0} x^{2} \sin\left(\frac{1}{x}\right) = 0 = f(0)$$ Step 2: Compute derivative at 0 using definition: $$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^{2} \sin\left(\frac{1}{h}\right)}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right) = 0$$ Step 3: Since limit exists, \(f\) is differentiable at 0 with \(f'(0) = 0\). 9. **Test differentiability of** \(f(x) = \begin{cases} \frac{x}{1 + e^{1/x}}, & x \neq 0 \\ 0, & x = 0 \end{cases}\) **at** \(x = 0\). Step 1: Check continuity at 0: $$\lim_{x \to 0} \frac{x}{1 + e^{1/x}} = 0 = f(0)$$ Step 2: Compute derivative at 0: $$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{\frac{h}{1 + e^{1/h}}}{h} = \lim_{h \to 0} \frac{1}{1 + e^{1/h}}$$ Step 3: The limit from right \(h \to 0^{+}\) is \(\frac{1}{1 + \infty} = 0\), from left \(h \to 0^{-}\) is \(\frac{1}{1 + 0} = 1\). Step 4: Since left and right limits differ, \(f'(0)\) does not exist; function is not differentiable at 0. 10. **Show** \(f(x) = |x - 3|\) **is continuous but not differentiable at** \(x = 3\). Step 1: Continuity: $$\lim_{x \to 3} |x - 3| = 0 = f(3)$$ Step 2: Left derivative: $$\lim_{h \to 0^{-}} \frac{|3 + h - 3| - 0}{h} = \lim_{h \to 0^{-}} \frac{-h}{h} = -1$$ Step 3: Right derivative: $$\lim_{h \to 0^{+}} \frac{|3 + h - 3| - 0}{h} = \lim_{h \to 0^{+}} \frac{h}{h} = 1$$ Step 4: Since left and right derivatives differ, not differentiable at 3. 11. **Show** \(f(x) = \begin{cases} 1 - x, & 0 < x < 1 \\ x^{2} - 1, & x \geq 1 \end{cases}\) **is continuous but not differentiable at** \(x = 1\). Step 1: Check continuity: $$\lim_{x \to 1^{-}} (1 - x) = 0$$ $$\lim_{x \to 1^{+}} (x^{2} - 1) = 0$$ $$f(1) = 1^{2} - 1 = 0$$ Step 2: Left derivative: $$\lim_{h \to 0^{-}} \frac{(1 - (1 + h)) - 0}{h} = \lim_{h \to 0^{-}} \frac{-h}{h} = -1$$ Step 3: Right derivative: $$\lim_{h \to 0^{+}} \frac{((1 + h)^{2} - 1) - 0}{h} = \lim_{h \to 0^{+}} \frac{2h + h^{2}}{h} = 2$$ Step 4: Since derivatives differ, not differentiable at 1. 12. **Show** \(f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}\) **is continuous but not differentiable at** \(x = 0\). Step 1: Continuity: $$\lim_{x \to 0} x \sin \frac{1}{x} = 0 = f(0)$$ Step 2: Derivative at 0: $$f'(0) = \lim_{h \to 0} \frac{h \sin \frac{1}{h} - 0}{h} = \lim_{h \to 0} \sin \frac{1}{h}$$ Step 3: Limit does not exist due to oscillation; hence not differentiable at 0. 13. **Calculate derivatives:** (i) \(y = \frac{ax^{2} + bx + c}{\sqrt{x}} = (ax^{2} + bx + c) x^{-1/2}\) Use product rule: $$y' = (2ax + b) x^{-1/2} + (ax^{2} + bx + c)(-\frac{1}{2} x^{-3/2})$$ (ii) \(y = 1 - \tan x\) $$y' = -\sec^{2} x$$ (iii) \(y = \frac{\sin x - \cos x}{\sin x + \cos x}\) Use quotient rule: $$y' = \frac{(\cos x + \sin x)(\sin x + \cos x) - (\sin x - \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^{2}}$$ (iv) \(y = x^{3} (\sec x + \csc x) e^{x} (\cos x + \sin x)\) Use product rule multiple times; derivative is lengthy. 14. **Determine** \(\frac{dy}{dx}\): (i) \(y = (2x + 5)^{5}\) $$\frac{dy}{dx} = 5(2x + 5)^{4} \times 2 = 10(2x + 5)^{4}$$ (ii) \(y = \cot^{-1}(\sin x)\) $$\frac{dy}{dx} = -\frac{\cos x}{1 + \sin^{2} x}$$ (iii) \(y = \sqrt{\cos(\log x)} = (\cos(\log x))^{1/2}\) $$\frac{dy}{dx} = \frac{1}{2} (\cos(\log x))^{-1/2} \times (-\sin(\log x)) \times \frac{1}{x} = -\frac{\sin(\log x)}{2x \sqrt{\cos(\log x)}}$$ (iv) \(y = x^{\sin x} + x^{x}\) Use logarithmic differentiation: $$\frac{d}{dx} x^{\sin x} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right)$$ $$\frac{d}{dx} x^{x} = x^{x} (\ln x + 1)$$ Sum gives \(\frac{dy}{dx}\). 15. **If** \(x^{2} + y^{2} = 4xy\), **find** \(\frac{dy}{dx}\). Step 1: Differentiate both sides implicitly: $$2x + 2y \frac{dy}{dx} = 4y + 4x \frac{dy}{dx}$$ Step 2: Rearrange terms: $$2y \frac{dy}{dx} - 4x \frac{dy}{dx} = 4y - 2x$$ $$\frac{dy}{dx} (2y - 4x) = 4y - 2x$$ Step 3: Solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx} = \frac{4y - 2x}{2y - 4x} = \frac{2(2y - x)}{2(y - 2x)} = \frac{2y - x}{y - 2x}$$