1. **Statement of the problem:** We are given the function $$g(x) = 2 - x - \ln{(x-1)^2}$$ defined on the domain $$]1, +\infty[$$. We need to calculate the limits $$\lim_{x \to +\infty} g(x)$$ and $$\lim_{x \to 1^+} g(x)$$. 2. **Recall the formulas and rules:** - The natural logarithm function $$\ln(x)$$ tends to $$+\infty$$ as $$x \to +\infty$$ and tends to $$-\infty$$ as $$x \to 0^+$$. - For limits involving logarithms and polynomials, analyze dominant terms. 3. **Calculate $$\lim_{x \to +\infty} g(x)$$:** $$g(x) = 2 - x - \ln{(x-1)^2} = 2 - x - 2\ln(x-1)$$ As $$x \to +\infty$$, $$x$$ grows without bound and $$\ln(x-1)$$ also tends to $$+\infty$$ but slower than $$x$$. So, $$\lim_{x \to +\infty} g(x) = \lim_{x \to +\infty} (2 - x - 2\ln(x-1)) = -\infty$$ because the linear term $$-x$$ dominates. 4. **Calculate $$\lim_{x \to 1^+} g(x)$$:** As $$x \to 1^+$$, $$x-1 \to 0^+$$, so $$\ln{(x-1)^2} = 2 \ln(x-1) \to 2 \times (-\infty) = -\infty$$ Thus, $$g(x) = 2 - x - \ln{(x-1)^2} = 2 - x - 2\ln(x-1)$$ Since $$-2\ln(x-1) \to +\infty$$, and $$2 - x \to 1$$, Therefore, $$\lim_{x \to 1^+} g(x) = +\infty$$. **Final answers:** $$\lim_{x \to +\infty} g(x) = -\infty$$ $$\lim_{x \to 1^+} g(x) = +\infty$$