1. **Evaluate** $$\lim_{x \to 0} \frac{a^x - b^x}{x}$$ Step 1: Recall the exponential limit property: $$\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$$. Step 2: Rewrite the expression: $$\frac{a^x - b^x}{x} = \frac{a^x - 1 + 1 - b^x}{x} = \frac{a^x - 1}{x} - \frac{b^x - 1}{x}$$ Step 3: Take the limit using the property: $$\lim_{x \to 0} \frac{a^x - b^x}{x} = \lim_{x \to 0} \frac{a^x - 1}{x} - \lim_{x \to 0} \frac{b^x - 1}{x} = \ln a - \ln b = \ln \left(\frac{a}{b}\right)$$ **Answer:** $$\boxed{\ln \left(\frac{a}{b}\right)}$$ 2. (i) **Calculate** $$\lim_{x \to \infty} \frac{2x^5 - 3x^2 + 5}{3x^5 + 2x^2 + 8}$$ Step 1: Divide numerator and denominator by $$x^5$$ (highest power): $$\frac{2 - 3x^{-3} + 5x^{-5}}{3 + 2x^{-3} + 8x^{-5}}$$ Step 2: As $$x \to \infty$$, terms with negative powers go to zero: $$\lim_{x \to \infty} = \frac{2}{3}$$ (ii) **Calculate** $$\lim_{x \to -1} \frac{3x^2 - 4x + 1}{x^2 - 4x + 3}$$ Step 1: Substitute $$x = -1$$: Numerator: $$3(-1)^2 - 4(-1) + 1 = 3 + 4 + 1 = 8$$ Denominator: $$(-1)^2 - 4(-1) + 3 = 1 + 4 + 3 = 8$$ Step 2: Limit is $$\frac{8}{8} = 1$$ **Answer:** (i) $$\frac{2}{3}$$, (ii) $$1$$ 3. **Show** $$\lim_{x \to 2} \frac{|x - 2|}{x - 2}$$ does not exist. Step 1: Consider left-hand limit (LHL) as $$x \to 2^-$$: $$\frac{|x - 2|}{x - 2} = \frac{-(x - 2)}{x - 2} = -1$$ Step 2: Consider right-hand limit (RHL) as $$x \to 2^+$$: $$\frac{|x - 2|}{x - 2} = \frac{x - 2}{x - 2} = 1$$ Step 3: Since LHL $$\neq$$ RHL, limit does not exist. **Answer:** Limit does not exist. 4. **Evaluate** $$\lim_{x \to 2} \frac{\sqrt{3 - x} - 1}{2 - x}$$ Step 1: Substitute $$x=2$$ directly: Numerator: $$\sqrt{3 - 2} - 1 = 1 - 1 = 0$$ Denominator: $$2 - 2 = 0$$ Step 2: Use conjugate to simplify: $$\frac{\sqrt{3 - x} - 1}{2 - x} \times \frac{\sqrt{3 - x} + 1}{\sqrt{3 - x} + 1} = \frac{3 - x - 1}{(2 - x)(\sqrt{3 - x} + 1)} = \frac{2 - x}{(2 - x)(\sqrt{3 - x} + 1)}$$ Step 3: Cancel $$2 - x$$: $$\frac{1}{\sqrt{3 - x} + 1}$$ Step 4: Substitute $$x=2$$: $$\frac{1}{\sqrt{1} + 1} = \frac{1}{2}$$ **Answer:** $$\boxed{\frac{1}{2}}$$ 5. **Examine continuity of** $$f(x) = x^3$$ at $$x=2$$ Step 1: Check $$f(2) = 2^3 = 8$$ Step 2: Calculate $$\lim_{x \to 2} f(x) = \lim_{x \to 2} x^3 = 8$$ Step 3: Since $$\lim_{x \to 2} f(x) = f(2)$$, function is continuous at $$x=2$$. **Answer:** 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 limit $$\neq$$ right limit, function is not continuous at $$x=0$$. **Answer:** Discontinuous 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: Left limit and value at 2: $$f(2) = k(2) + 5 = 2k + 5$$ Step 2: Right limit: $$\lim_{x \to 2^+} f(x) = 2 - 1 = 1$$ Step 3: For continuity: $$2k + 5 = 1 \implies 2k = -4 \implies k = -2$$ **Answer:** $$k = -2$$ 8. **Show differentiability of** $$f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0 \end{cases}$$ at $$x=0$$ Step 1: Check continuity at 0: $$\lim_{x \to 0} x^2 \sin \frac{1}{x} = 0 = f(0)$$ Step 2: Compute derivative at 0 using definition: $$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin \frac{1}{h}}{h} = \lim_{h \to 0} h \sin \frac{1}{h}$$ Step 3: Since $$|h \sin \frac{1}{h}| \leq |h|$$ and $$\lim_{h \to 0} h = 0$$, derivative is 0. **Answer:** Differentiable at $$x=0$$ 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: $$\lim_{x \to 0} \frac{x}{1 + e^{1/x}}$$ As $$x \to 0^+$$, $$e^{1/x} \to \infty$$, so denominator $$\to \infty$$, limit $$\to 0$$. As $$x \to 0^-$$, $$e^{1/x} \to 0$$, so denominator $$\to 1$$, limit $$\to 0$$. Step 2: Compute derivative at 0: $$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{\frac{h}{1 + e^{1/h}} - 0}{h} = \lim_{h \to 0} \frac{1}{1 + e^{1/h}}$$ Step 3: Left and right limits differ: Right limit $$= 0$$, left limit $$= 1$$. Step 4: Derivative does not exist at 0. **Answer:** Not differentiable at $$x=0$$. 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: Derivative from left: $$\lim_{h \to 0^-} \frac{|3 + h - 3| - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$$ Step 3: Derivative from right: $$\lim_{h \to 0^+} \frac{h}{h} = 1$$ Step 4: Since left and right derivatives differ, not differentiable at 3. **Answer:** Continuous but not differentiable at $$x=3$$. 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: $$\lim_{x \to 1^-} f(x) = 1 - 1 = 0$$ $$\lim_{x \to 1^+} f(x) = 1^2 - 1 = 0$$ $$f(1) = 0$$ Step 2: Derivative from left: $$\lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^-} \frac{1 - (1 + h) - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$$ Step 3: Derivative from right: $$\lim_{h \to 0^+} \frac{(1 + h)^2 - 1 - 0}{h} = \lim_{h \to 0^+} \frac{1 + 2h + h^2 - 1}{h} = \lim_{h \to 0^+} (2 + h) = 2$$ Step 4: Since derivatives differ, not differentiable at 1. **Answer:** Continuous but not differentiable at $$x=1$$. 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 because $$\sin \frac{1}{h}$$ oscillates. **Answer:** Continuous but not differentiable at $$x=0$$. 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) $$y = \frac{1 - \tan x}{\tan x} = \frac{1}{\tan x} - 1 = \cot x - 1$$ Derivative: $$y' = -\csc^2 x$$ (iii) $$y = \frac{\sin x - \cos x}{\sin x + \cos x}$$ Use quotient rule: $$y' = \frac{(\cos x + \sin x)(\sin x + \cos x) - (\sin x - \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2}$$ Simplify numerator: $$= \frac{(\cos x + \sin x)^2 - (\sin x - \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2}$$ (iv) $$y = x^3 (\sec x + \csc x)$$ Use product rule: $$y' = 3x^2 (\sec x + \csc x) + x^3 (\sec x \tan x - \csc x \cot x)$$ (v) $$y = e^x (\cos x + \sin x)$$ Use product rule: $$y' = e^x (\cos x + \sin x) + e^x (-\sin x + \cos x) = e^x (2 \cos x)$$ 14. **Determine** $$\frac{dy}{dx}$$ for: (i) $$y = (2x + 5)^5$$ $$\frac{dy}{dx} = 5(2x + 5)^4 \times 2 = 10(2x + 5)^4$$ (ii) $$y = \cot^{-1}(\sin x)$$ $$\frac{dy}{dx} = -\frac{1}{1 + (\sin x)^2} \cdot \cos x = -\frac{\cos x}{1 + \sin^2 x}$$ (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) $$y = x^{\sin x} + x^x$$ For $$x^{\sin x}$$: $$\frac{d}{dx} x^{\sin x} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right)$$ For $$x^x$$: $$\frac{d}{dx} x^x = x^x (\ln x + 1)$$ So, $$\frac{dy}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right) + x^x (\ln x + 1)$$ 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: $$2y \frac{dy}{dx} - 4x \frac{dy}{dx} = 4y - 2x$$ Step 3: Factor $$\frac{dy}{dx}$$: $$\frac{dy}{dx} (2y - 4x) = 4y - 2x$$ Step 4: Solve for $$\frac{dy}{dx}$$: $$\frac{dy}{dx} = \frac{4y - 2x}{2y - 4x} = \frac{2(2y - x)}{2(y - 2x)} = \frac{2y - x}{y - 2x}$$ **Answer:** $$\boxed{\frac{dy}{dx} = \frac{2y - x}{y - 2x}}$$