1) Calculate the limit $\lim_{x \to +\infty} \left[\sqrt{x^3 + 5} - x\sqrt{x}\right]$. Step 1. State the problem: Find $\lim_{x \to +\infty} \left[\sqrt{x^3 + 5} - x\sqrt{x}\right]$. Step 2. Use the fact that $\sqrt{x^3 + 5} = \sqrt{x^3(1 + \frac{5}{x^3})} = x^{3/2}\sqrt{1 + \frac{5}{x^3}}$. Step 3. Rewrite the expression: $$\sqrt{x^3 + 5} - x\sqrt{x} = x^{3/2}\sqrt{1 + \frac{5}{x^3}} - x^{3/2} = x^{3/2}\left(\sqrt{1 + \frac{5}{x^3}} - 1\right)$$ Step 4. For large $x$, use the binomial approximation $\sqrt{1 + t} \approx 1 + \frac{t}{2}$ when $t \to 0$: $$\sqrt{1 + \frac{5}{x^3}} \approx 1 + \frac{5}{2x^3}$$ Step 5. Substitute approximation: $$x^{3/2} \left(1 + \frac{5}{2x^3} - 1\right) = x^{3/2} \cdot \frac{5}{2x^3} = \frac{5}{2} x^{3/2 - 3} = \frac{5}{2} x^{-3/2}$$ Step 6. As $x \to +\infty$, $x^{-3/2} \to 0$, so the limit is: $$\boxed{0}$$ --- 2) Calculate the limit $\lim_{x \to 1} \frac{x^2 - 2x + 1}{x^3 - 3x^2 + 3x - 1}$. Step 1. Factor numerator and denominator: - Numerator: $x^2 - 2x + 1 = (x - 1)^2$ - Denominator: Recognize $x^3 - 3x^2 + 3x - 1 = (x - 1)^3$ (binomial expansion of $(x-1)^3$) Step 2. Rewrite the limit: $$\lim_{x \to 1} \frac{(x - 1)^2}{(x - 1)^3} = \lim_{x \to 1} \frac{1}{x - 1}$$ Step 3. As $x \to 1$, $\frac{1}{x - 1}$ tends to $\pm \infty$ depending on the direction. Step 4. Since the limit from the left and right differ, the limit does not exist (infinite discontinuity). Answer: $$\boxed{\text{The limit does not exist (infinite).}}$$ --- 3) Calculate the limit $\lim_{x \to +\infty} \frac{\sqrt{3x + 1} - 5}{7x + 1}$. Step 1. For large $x$, numerator behaves like $\sqrt{3x} - 5 = \sqrt{3} \sqrt{x} - 5$. Step 2. Denominator behaves like $7x$. Step 3. The numerator grows like $\sqrt{x}$, denominator grows like $x$. Step 4. Since $\sqrt{x} / x = 1/\sqrt{x} \to 0$, the whole fraction tends to 0. Answer: $$\boxed{0}$$ --- 4) Calculate the derivative of $y = (3x + 5)\sqrt{2 - x}$. Step 1. Use product rule: $y' = u'v + uv'$ where $u = 3x + 5$, $v = (2 - x)^{1/2}$. Step 2. Compute derivatives: - $u' = 3$ - $v' = \frac{1}{2}(2 - x)^{-1/2} \cdot (-1) = -\frac{1}{2\sqrt{2 - x}}$ Step 3. Substitute: $$y' = 3 \cdot \sqrt{2 - x} + (3x + 5) \cdot \left(-\frac{1}{2\sqrt{2 - x}}\right) = 3\sqrt{2 - x} - \frac{3x + 5}{2\sqrt{2 - x}}$$ Answer: $$\boxed{y' = 3\sqrt{2 - x} - \frac{3x + 5}{2\sqrt{2 - x}}}$$ --- 5) Calculate the derivative of $y = \frac{(2x - 1)^3}{x + 5}$. Step 1. Use quotient rule: $$y' = \frac{(u'v - uv')}{v^2}$$ where $u = (2x - 1)^3$, $v = x + 5$. Step 2. Compute derivatives: - $u' = 3(2x - 1)^2 \cdot 2 = 6(2x - 1)^2$ - $v' = 1$ Step 3. Substitute: $$y' = \frac{6(2x - 1)^2 (x + 5) - (2x - 1)^3 \cdot 1}{(x + 5)^2} = \frac{(2x - 1)^2 \left[6(x + 5) - (2x - 1)\right]}{(x + 5)^2}$$ Step 4. Simplify inside bracket: $$6(x + 5) - (2x - 1) = 6x + 30 - 2x + 1 = 4x + 31$$ Answer: $$\boxed{y' = \frac{(2x - 1)^2 (4x + 31)}{(x + 5)^2}}$$ --- 6) Calculate the derivative of $y = x^2 - x^{-2} + x^{\frac{5}{3}} - x^{-\frac{1}{4}}$. Step 1. Use power rule $\frac{d}{dx} x^n = n x^{n-1}$: - $\frac{d}{dx} x^2 = 2x$ - $\frac{d}{dx} x^{-2} = -2 x^{-3}$ - $\frac{d}{dx} x^{5/3} = \frac{5}{3} x^{2/3}$ - $\frac{d}{dx} x^{-1/4} = -\frac{1}{4} x^{-5/4}$ Step 2. Combine: $$y' = 2x + 2x^{-3} + \frac{5}{3} x^{2/3} + \frac{1}{4} x^{-5/4}$$ Answer: $$\boxed{y' = 2x + 2x^{-3} + \frac{5}{3} x^{2/3} + \frac{1}{4} x^{-5/4}}$$ --- 7) Given $f(x) = \sqrt{x - 2} - \sqrt{x + 1}$, determine the domain. Step 1. The expressions under square roots must be $\geq 0$: - $x - 2 \geq 0 \Rightarrow x \geq 2$ - $x + 1 \geq 0 \Rightarrow x \geq -1$ Step 2. Domain is intersection: $$\boxed{[2, +\infty)}$$ --- 8) Calculate the limits of $f(x)$ at domain boundaries. Step 1. At $x \to 2^+$: $$f(2) = \sqrt{0} - \sqrt{3} = -\sqrt{3}$$ Step 2. At $x \to +\infty$: $$f(x) = \sqrt{x - 2} - \sqrt{x + 1} = \sqrt{x}\left(\sqrt{1 - \frac{2}{x}} - \sqrt{1 + \frac{1}{x}}\right)$$ Step 3. Use binomial approximation: $$\sqrt{1 - \frac{2}{x}} \approx 1 - \frac{1}{x}, \quad \sqrt{1 + \frac{1}{x}} \approx 1 + \frac{1}{2x}$$ Step 4. Difference inside parentheses: $$\left(1 - \frac{1}{x}\right) - \left(1 + \frac{1}{2x}\right) = -\frac{3}{2x}$$ Step 5. Multiply by $\sqrt{x} = x^{1/2}$: $$x^{1/2} \cdot \left(-\frac{3}{2x}\right) = -\frac{3}{2} x^{-1/2} \to 0$$ Answer: $$\boxed{\lim_{x \to 2^+} f(x) = -\sqrt{3}, \quad \lim_{x \to +\infty} f(x) = 0}$$ --- 9) Calculate the derivative of $f(x) = \sqrt{x - 2} - \sqrt{x + 1}$. Step 1. Use derivative of $\sqrt{u} = \frac{1}{2\sqrt{u}} u'$: - $\frac{d}{dx} \sqrt{x - 2} = \frac{1}{2\sqrt{x - 2}}$ - $\frac{d}{dx} \sqrt{x + 1} = \frac{1}{2\sqrt{x + 1}}$ Step 2. Combine: $$f'(x) = \frac{1}{2\sqrt{x - 2}} - \frac{1}{2\sqrt{x + 1}}$$ Answer: $$\boxed{f'(x) = \frac{1}{2\sqrt{x - 2}} - \frac{1}{2\sqrt{x + 1}}}$$ --- 10) Consider $f(x) = 2x^2 - 3x + 1$. Step 1. Domain: polynomial functions are defined for all real numbers. $$\boxed{(-\infty, +\infty)}$$ Step 2. Limits at boundaries: - $\lim_{x \to -\infty} f(x) = +\infty$ (since leading term $2x^2$ dominates) - $\lim_{x \to +\infty} f(x) = +\infty$ Step 3. Derivative: $$f'(x) = 4x - 3$$ Answer: $$\boxed{\text{Domain} = (-\infty, +\infty), \quad \lim_{x \to \pm \infty} f(x) = +\infty, \quad f'(x) = 4x - 3}$$