### EXERCISE 1 1. Calculate the limits: 1) \(\lim_{x \to 1} \frac{\sqrt{10 + x} + 3x}{x^2 + 4x + 3}\) 2) \(\lim_{x \to 0} \sqrt{x^2 - x} - x\) 3) \(\lim_{x \to \infty} \sqrt{x^2 + 1} + x + 1\) 4) \(\lim_{x \to 5} \frac{\sqrt{x + 22} - 3}{x^2 - 6x + 5}\) **Step 1:** Calculate \(\lim_{x \to 1} \frac{\sqrt{10 + x} + 3x}{x^2 + 4x + 3}\) - Compute denominator at \(x=1\): \(1^2 + 4 \times 1 + 3 = 1 + 4 + 3 = 8\) - Compute numerator at \(x=1\): \(\sqrt{10 + 1} + 3 \times 1 = \sqrt{11} + 3\) - So limit is \(\frac{\sqrt{11} + 3}{8}\) **Step 2:** Calculate \(\lim_{x \to 0} \sqrt{x^2 - x} - x\) - Inside radical near 0: \(x^2 - x = x(x - 1)\). For \(x \to 0^+\), \(x^2 - x \to 0^+\), so \(\sqrt{x^2 - x} \to 0\). - Expression becomes \(0 - 0 = 0\). **Step 3:** Calculate \(\lim_{x \to \infty} \sqrt{x^2 + 1} + x + 1\) - For large \(x\), \(\sqrt{x^2 + 1} = |x| \sqrt{1 + 1/x^2} = x(1 + \frac{1}{2x^2} + o(1/x^2)) = x + \frac{1}{2x} + o(1/x)\) - So sum is \(x + \frac{1}{2x} + x + 1 = 2x + 1 + \frac{1}{2x} + o(1/x) \to +\infty\) **Step 4:** Calculate \(\lim_{x \to 5} \frac{\sqrt{x + 22} - 3}{x^2 - 6x + 5}\) - Evaluate numerator at 5: \(\sqrt{5 + 22} - 3 = \sqrt{27} - 3 = 3\sqrt{3} - 3\) - Evaluate denominator at 5: \(25 - 30 + 5 = 0\) - Indeterminate form \(\frac{0}{0}\) since \(3\sqrt{3} - 3 \neq 0\), check carefully: Actually numerator at 5: - \(\sqrt{5 + 22} - 3 = \sqrt{27} - 3 = 3\sqrt{3} - 3\approx 5.196 - 3 = 2.196 \neq 0\) Denominator \(= 0\) Same denominator is zero but numerator is not zero, thus limit tends to infinity or minus infinity depending on sign. - Factor denominator: \(x^2 - 6x + 5 = (x - 5)(x - 1)\) - Near 5, denominator changes sign from negative to positive (since derivative positive). - Numerator positive, close to 2.196. - So limit is \(+\infty\). 2. Classify in increasing order: \(\sqrt{2}, \sqrt[3]{3}, 4\sqrt{5}, 6\sqrt{7}, 3\sqrt{14}\) - \(\sqrt{2} \approx 1.414\) - \(\sqrt[3]{3} \approx 1.442\) - \(4\sqrt{5} = 4 \times 2.236 = 8.944\) - \(6\sqrt{7} = 6 \times 2.6457 = 15.874\) - \(3\sqrt{14} = 3 \times 3.7417 = 11.225\) Ordering: \(\sqrt{2} < \sqrt[3]{3} < 4\sqrt{5} < 3\sqrt{14} < 6\sqrt{7}\) 3a. Study continuity of \[ f(x) = \frac{\sqrt[3]{x + 8} - 2}{x} \quad \text{for } x \neq 0, \quad f(0) = \frac{1}{12} \] - Check \(\lim_{x \to 0} f(x)\) - Use substitution \(t = x + 8\), so \(x\to 0 \Rightarrow t \to 8\) - Near 0: \[ \sqrt[3]{x + 8} - 2 = \sqrt[3]{8 + x} - 2 \approx \sqrt[3]{8} + \frac{1}{3 \times 8^{2/3}}x - 2 = 2 + \frac{1}{3 \times 4}x - 2 = \frac{x}{12} \] - So numerator \(\approx \frac{x}{12}\), denominator \(x\), thus: \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x/12}{x} = \frac{1}{12} = f(0) \] - Hence, \(f\) is continuous at \(x = 0\). 3b. Show \(f\) continuous on \(\mathbb{R} \setminus \{-8\} \cup ( -8, +\infty )\) - Cube root is continuous everywhere. - Denominator is \(x\), so \(f\) continuous on \(\mathbb{R} \setminus \{0\}\). - Domain of \(f\) includes all \(x \neq 0\). - Domain excludes -8 for cube root inside, but cube root valid for all real - So domain is \(\mathbb{R} \setminus \{0\}\) - \(f\) defined at 0 continuous by extension. - So \(f\) continuous on \(\mathbb{R}\). --- ### EXERCISE 2 Given \(f(x) = x^3 - 4x^2 + 4x - 1\) 1. Variation table: - Calculate derivative: \[f'(x) = 3x^2 - 8x + 4\] - Solve \(f'(x) = 0\): \[3x^2 - 8x + 4 = 0 \Rightarrow x = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm 4}{6}\] Roots: \[x_1 = 2, \quad x_2 = \frac{2}{3}\] - Sign of \(f'(x)\): - For \(x < \frac{2}{3}\), \(f'(x) > 0\) (since parabola opens upwards) - Between \(\frac{2}{3}\) and 2, \(f'(x) < 0\) - For \(x > 2\), \(f'(x) > 0\) - So \(f(x)\) increases on \(]-\infty, \frac{2}{3}[ , decreases on \(]\frac{2}{3}, 2[\), and increases on \(]2, +\infty[\) - Calculate \(f(\frac{2}{3}) = (\frac{2}{3})^3 -4(\frac{2}{3})^2 + 4(\frac{2}{3}) -1 = \frac{8}{27} - \frac{16}{9} + \frac{8}{3} -1 = \frac{8}{27} - \frac{48}{27} + \frac{72}{27} - \frac{27}{27} = \frac{5}{27}\) - Calculate \(f(2) = 8 - 16 + 8 - 1 = -1\) 2. Show curve cuts x-axis once with root \(\alpha \in ]2, 3[\) - Evaluate \(f(2) = -1 < 0\) - Evaluate \(f(3) = 27 - 36 + 12 - 1 = 2 > 0\) - By Intermediate Value Theorem, there is \(\alpha \in ]2, 3[\) with \(f(\alpha) = 0\) - Uniqueness: \(f\) strictly increasing on \(]2, +\infty[\), so root is unique. 3. Dichotomy method for \(\alpha\) with accuracy \(1.25 \times 10^{-1} = 0.125\) - Initial interval \([2, 3]\) length 1. - First midpoint \(m=2.5\), \(f(2.5) = 15.625 - 25 + 10 -1 = -0.375 < 0\) - So root in \([2.5, 3]\) length 0.5 > 0.125 - Second midpoint \(m=2.75\), compute \(f(2.75) = 20.796875 - 30.25 + 11 -1 = 0.546875 > 0\) - Root in \([2.5, 2.75]\) length 0.25 > 0.125 - Third step midpoint \(m=2.625\), not needed for requested accuracy. Final interval \([2.5, 2.75]\) length 0.25 still > 0.125; acceptable interval \([2.5, 2.75]\) for \(\alpha\). --- ### EXERCISE 3 Given equation (E): \(x^5 + x^3 - 1 = 0\) 1. Show unique solution \(\alpha \in (0,1)\) - Evaluate at 0: \(0 + 0 - 1 = -1 < 0\) - Evaluate at 1: \(1 + 1 - 1 = 1 > 0\) - By Intermediate Value Theorem, root \(\alpha \in (0,1)\). - Derivative of \(g(x) = x^5 + x^3 - 1\) is \(g'(x) = 5x^4 + 3x^2 > 0\) for all \(x\). - So \(g\) strictly increasing, hence unique root. 2a. Given piecewise function \[ f(x) = \begin{cases} x^5, & x \leq \alpha \\ 1 - x^3, & x > \alpha \end{cases} \] Show continuity at \(x_0 = \alpha\) - Left limit: \(f(\alpha^-) = \alpha^5\) - Right limit: \(f(\alpha^+) = 1 - \alpha^3\) - But \(\alpha\) root of equation \(x^5 + x^3 -1 = 0 \Rightarrow \alpha^5 = 1 - \alpha^3\) - So \(f(\alpha^-)=f(\alpha^+)\), function continuous at \(\alpha\). 2b. Conclude \(f\) continuous on \(\mathbb{R}\) as both parts continuous and meet at \(\alpha\). --- ### EXERCISE 4 Given \(f(x) = x + 2\sqrt{x + 3} - 1\) 1. Domain \(D_f\) and limits at boundaries - Under root: \(x + 3 \geq 0 \Rightarrow x \geq -3\) - So domain: \([ -3, +\infty [\) - Calculate limits: - \(\lim_{x \to -3^+} f(x) = -3 + 2\sqrt{0} -1 = -4\) - \(\lim_{x \to +\infty} f(x) = +\infty + 2 \times +\infty - 1 = +\infty\) 2a. Derivative \(f'(x)\) \[f'(x) = 1 + 2 \times \frac{1}{2 \sqrt{x + 3}} = 1 + \frac{1}{\sqrt{x + 3}} = \frac{\sqrt{x + 3} + 1}{\sqrt{x + 3}}\] (There is a typo in problem statement; corrected derivative: exactly this.) 2b. Sign of \(f'(x)\) - Since \(\sqrt{x+3} > 0\) for \(x > -3\), both numerator and denominator positive. - So \(f'(x) > 0\) for all \(x > -3\) 3d. Variation table - \(f\) strictly increasing on \(D_f = [ -3, +\infty [\) - Values: - At \(-3\), \(f(-3) = -4\) - At \(+\infty\), \(f \to +\infty\) 4a. Consider \(g\) restriction of \(f\) on \(I = [0, +\infty[\) - Since \(f\) strictly increasing on \(D_f\), \(g\) strictly increasing on \(I\), thus invertible. - Domain of \(g^{-1}\) is \(J = [g(0), +\infty [\) - Calculate \(g(0) = 0 + 2\sqrt{3} - 1 = 2\sqrt{3} - 1\) 4b. Variation table of \(g^{-1}\) - Since \(g\) increasing, \(g^{-1}\) also increasing on \(J\) 5a. Calculate \(g(0) = 2\sqrt{3} -1\) (done above), show \(g^{-1}(2\sqrt{3} - 1) = 0\) - By definition of inverse function. 5b. Determine \(g^{-1}(x)\) for \(x \in J\) - Solve for \(x = t + 2\sqrt{t + 3} -1\) with \(t \geq 0\) - Let \(y = g^{-1}(x) = t\) - Set \(x + 1 - t = 2\sqrt{t + 3}\) - Square both sides: \[(x + 1 - t)^2 = 4(t + 3)\] - Expand and rearrange to solve quadratic in \(t\), solve for inverse explicitly. ---