\( \lim _{x \rightarrow +\infty } (x(\sqrt {\frac{x}{x+1}}-1)) \)

To evaluate the limit \[ \lim _{x \rightarrow +\infty } \left(x\left(\sqrt {\frac{x}{x+1}}-1\right)\right), \] we start by simplifying the expression inside the limit. First, we rewrite \(\sqrt{\frac{x}{x+1}}\): \[ \sqrt{\frac{x}{x+1}} = \sqrt{\frac{x}{x(1+\frac{1}{x})}} = \sqrt{\frac{1}{1+\frac{1}{x}}}. \] As \(x\) approaches infinity, the term \(\frac{1}{x}\) approaches \(0\). Therefore, \[ \sqrt{1+\frac{1}{x}} \to 1 \implies \sqrt{\frac{x}{x+1}} \to \sqrt{1} = 1. \] Next, let's rewrite the limit expression: \[ \sqrt{\frac{x}{x+1}} - 1 = \frac{\sqrt{\frac{x}{x+1}} - 1}{1} = \frac{\frac{x}{x+1} - 1}{\sqrt{\frac{x}{x+1}} + 1} = \frac{\frac{x - (x + 1)}{x + 1}}{\sqrt{\frac{x}{x+1}} + 1} = \frac{\frac{-1}{x + 1}}{\sqrt{\frac{x}{x+1}} + 1}. \] Now, we can multiply this form by \(x\): \[ x\left(\sqrt{\frac{x}{x+1}} - 1\right) = x \cdot \frac{-1}{(x + 1)(\sqrt{\frac{x}{x+1}} + 1)}. \] This leads to \[ \frac{-x}{(x + 1)(\sqrt{\frac{x}{x+1}} + 1)}. \] As \(x \to +\infty\): 1. The term \(x + 1 \sim x\). 2. The term \(\sqrt{\frac{x}{x+1}} \sim \sqrt{\frac{x}{x}} = 1\). 3. Therefore, \(\sqrt{\frac{x}{x+1}} + 1 \sim 2\). So we have: \[ \lim _{x \rightarrow +\infty } \frac{-x}{(x + 1)(\sqrt{\frac{x}{x+1}} + 1)} \sim \frac{-x}{x \cdot 2} = \frac{-1}{2}. \] Thus, the limit is: \[ \boxed{-\frac{1}{2}}. \]