\( \lim _ { x \rightarrow + \infty } x ^ { \alpha } ( \sqrt { x ^ { 2 } + 2 x + 3 } - x - 1 ) \)

To solve this limit, we start by simplifying the expression inside the limit. We can rewrite \(\sqrt{x^2 + 2x + 3}\) as follows: \[ \sqrt{x^2 + 2x + 3} = \sqrt{x^2(1 + \frac{2}{x} + \frac{3}{x^2})} = x\sqrt{1 + \frac{2}{x} + \frac{3}{x^2}} \] As \(x \rightarrow +\infty\), we can use the binomial expansion for \(\sqrt{1 + u}\) where \(u = \frac{2}{x} + \frac{3}{x^2}\): \[ \sqrt{1 + \frac{2}{x} + \frac{3}{x^2}} \approx 1 + \frac{1}{2}\left(\frac{2}{x} + \frac{3}{x^2}\right) \] This simplifies to: \[ 1 + \frac{1}{x} + \frac{3/2}{x^2} + O\left(\frac{1}{x^3}\right) \] Substituting this back, we have: \[ \sqrt{x^2 + 2x + 3} \approx x\left(1 + \frac{1}{x} + \frac{3/2}{x^2}\right) = x + 1 + \frac{3/2}{x} + O\left(\frac{1}{x^2}\right) \] Now substituting this approximation into our original expression we find: \[ \sqrt{x^2 + 2x + 3} - x - 1 \approx \left(x + 1 + \frac{3/2}{x} + O\left(\frac{1}{x^2}\right)\right) - x - 1 = \frac{3/2}{x} + O\left(\frac{1}{x^2}\right) \] Thus, we can rewrite the limit as: \[ \lim_{x \rightarrow +\infty} x^{\alpha} \left(\frac{3/2}{x} + O\left(\frac{1}{x^2}\right)\right) = \lim_{x \rightarrow +\infty} \left(\frac{3/2}{x^{1-\alpha}} + O\left(\frac{1}{x^{2-\alpha}}\right)\right) \] Now we analyze the limit based on the value of \(\alpha\): 1. If \(\alpha < 1\), \(\frac{3/2}{x^{1-\alpha}} \rightarrow +\infty\). 2. If \(\alpha = 1\), \(\frac{3/2}{x^{1-\alpha}} \rightarrow \frac{3}{2}\). 3. If \(\alpha > 1\), \(\frac{3/2}{x^{1-\alpha}} \rightarrow 0\). Therefore, we conclude: \[ \lim_{x \rightarrow +\infty} x^{\alpha} \left(\sqrt{x^2 + 2x + 3} - x - 1\right) = \begin{cases} +\infty & \text{if } \alpha < 1 \\ \frac{3}{2} & \text{if } \alpha = 1 \\ 0 & \text{if } \alpha > 1 \end{cases} \]