g) \( \lim _{x \rightarrow+\infty} \frac{\sqrt{x^{2}+2 x+4}-x}{x-\sqrt{x^{2}-x+1}}, \mathrm{R}: 2 \)

To solve the limit \[ \lim_{x \rightarrow+\infty} \frac{\sqrt{x^{2}+2x+4} - x}{x - \sqrt{x^{2}-x+1}}, \] we will first simplify both the numerator and the denominator. **Step 1: Simplify the numerator** For the numerator \(\sqrt{x^{2}+2x+4} - x\): \[ \sqrt{x^{2}+2x+4} = \sqrt{x^2(1 + \frac{2}{x} + \frac{4}{x^2})} = x\sqrt{1 + \frac{2}{x} + \frac{4}{x^2}}. \] Thus, \[ \sqrt{x^{2}+2x+4} - x = x\left(\sqrt{1 + \frac{2}{x} + \frac{4}{x^2}} - 1\right). \] Using the binomial expansion, we have \(\sqrt{1 + u} \approx 1 + \frac{u}{2}\) for \(u\) close to 0: \[ \sqrt{1 + \frac{2}{x} + \frac{4}{x^2}} \approx 1 + \frac{1}{x} + \frac{2}{x^2}. \] So, \[ \sqrt{1 + \frac{2}{x} + \frac{4}{x^2}} - 1 \approx \frac{1}{x} + \frac{2}{x^2}. \] Thus, \[ \sqrt{x^{2}+2x+4} - x \approx x\left(\frac{1}{x} + \frac{2}{x^2}\right) = 1 + \frac{2}{x}. \] **Step 2: Simplify the denominator** For the denominator \(x - \sqrt{x^{2} - x + 1}\): \[ \sqrt{x^{2} - x + 1} = \sqrt{x^{2}\left(1 - \frac{1}{x} + \frac{1}{x^2}\right)} = x\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}}. \] Similarly, expanding \(\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}}\): \[ \sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} \approx 1 - \frac{1}{2x} + O\left(\frac{1}{x^2}\right). \] Thus, \[ x - \sqrt{x^{2} - x + 1} \approx x\left(1 - \left(1 - \frac{1}{2x} + O\left(\frac{1}{x^2}\right)\right)\right) = \frac{1}{2} + O\left(\frac{1}{x}\right). \] **Step 3: Combine the simplifications in the limit** Now substituting back into the limit: \[ \lim_{x \to +\infty} \frac{1 + \frac{2}{x}}{\frac{1}{2} + O\left(\frac{1}{x}\right)}. \] As \(x\) approaches infinity, we get \[ \lim_{x \to +\infty} \frac{1}{\frac{1}{2}} = 2. \] Therefore, the final result is \[ \boxed{2}. \]