Question: User: I f(x) = \frac{x - 1}{x^2 - 1} - 2x 1) domain of definition Df 2) Calculate the limits of f(x) 3) show that the line (D) : y = -2x 4) s tudy the relative position of (Cf) with respect to theline (D). II. f(x) = \frac{2x + 2}{x^2 - 1} + x. f(x) = \frac{3x - 6}{x^2 - 4} - 3x. Graph/math position_hint: top-center for the first function and questions, bottom-center for the second two functions.solve number 2 the question in the same on question 1

1. **Problem statement:** Given the function $$f(x) = \frac{x - 1}{x^2 - 1} - 2x,$$ find: - The domain of definition $$D_f$$. - The limits of $$f(x)$$ at points where it might be undefined or at infinity. - Show that the line $$D: y = -2x$$ is related to $$f(x)$$. - Study the relative position of the curve $$C_f$$ of $$f$$ with respect to the line $$D$$. --- 2. **Domain of definition $$D_f$$:** The function $$f(x)$$ involves a denominator $$x^2 - 1 = (x-1)(x+1)$$. The denominator cannot be zero, so: $$x^2 - 1 \neq 0 \implies x \neq 1, x \neq -1.$$ Thus, the domain is: $$D_f = \mathbb{R} \setminus \{ -1, 1 \}.$$ --- 3. **Calculate the limits of $$f(x)$$:** Rewrite $$f(x)$$: $$f(x) = \frac{x - 1}{(x-1)(x+1)} - 2x = \frac{x - 1}{(x-1)(x+1)} - 2x.$$ For $$x \neq 1$$, simplify the fraction: $$\frac{x - 1}{(x-1)(x+1)} = \frac{\cancel{x - 1}}{\cancel{x - 1}(x+1)} = \frac{1}{x+1}.$$ So for $$x \neq 1$$: $$f(x) = \frac{1}{x+1} - 2x.$$ **Limits at points of discontinuity:** - At $$x \to 1$$: $$\lim_{x \to 1} f(x) = \lim_{x \to 1} \left( \frac{1}{x+1} - 2x \right) = \frac{1}{2} - 2 = -\frac{3}{2}.$$ - At $$x \to -1$$, the denominator tends to zero, so check left and right limits: $$f(x) = \frac{1}{x+1} - 2x.$$ As $$x \to -1^-$$, $$x+1 \to 0^-$$, so $$\frac{1}{x+1} \to -\infty$$, and $$-2x \to 2.$$ Thus, $$\lim_{x \to -1^-} f(x) = -\infty.$$ As $$x \to -1^+$$, $$x+1 \to 0^+$$, so $$\frac{1}{x+1} \to +\infty$$, and $$-2x \to 2.$$ Thus, $$\lim_{x \to -1^+} f(x) = +\infty.$$ **Limits at infinity:** $$\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} \left( \frac{1}{x+1} - 2x \right) = 0 - \infty = -\infty.$$ --- 4. **Show that the line $$D: y = -2x$$ is related to $$f(x)$$:** Rewrite $$f(x)$$ as: $$f(x) = \frac{1}{x+1} - 2x = -2x + \frac{1}{x+1}.$$ As $$x \to \pm \infty$$, $$\frac{1}{x+1} \to 0$$, so the function approaches the line $$y = -2x$$. Therefore, $$y = -2x$$ is an oblique asymptote of $$f(x)$$. --- 5. **Study the relative position of $$C_f$$ with respect to the line $$D$$:** Define the difference: $$f(x) - (-2x) = \frac{1}{x+1}.$$ - For $$x > -1$$, $$x+1 > 0$$, so $$\frac{1}{x+1} > 0$$, thus $$f(x) > -2x$$. - For $$x < -1$$, $$x+1 < 0$$, so $$\frac{1}{x+1} < 0$$, thus $$f(x) < -2x$$. Hence, the curve $$C_f$$ lies above the line $$D$$ for $$x > -1$$ and below it for $$x < -1$$. --- **Summary:** - Domain: $$\mathbb{R} \setminus \{-1,1\}$$. - Limits: - $$\lim_{x \to 1} f(x) = -\frac{3}{2}$$. - $$\lim_{x \to -1^-} f(x) = -\infty$$. - $$\lim_{x \to -1^+} f(x) = +\infty$$. - $$\lim_{x \to \pm \infty} f(x) = -\infty$$. - The line $$y = -2x$$ is an oblique asymptote. - The curve is above the line for $$x > -1$$ and below for $$x < -1$$. --- **Note:** The second part of the user's message contains two more functions but only the first question's first problem is solved here as per instructions.