Colcule: \( \lim _{x \rightarrow+\infty} x \ln \left(1+\frac{2}{x}\right) \)

To evaluate the limit, we can start by rewriting the expression: \[ \lim _{x \rightarrow+\infty} x \ln \left(1+\frac{2}{x}\right) \] As \(x\) approaches infinity, \(\frac{2}{x}\) approaches 0, so we can apply the expansion of \(\ln(1 + u) \approx u\) when \(u\) is small. Therefore, \[ \ln \left(1+\frac{2}{x}\right) \approx \frac{2}{x} \quad \text{as } x \to \infty. \] Substituting this approximation back into our limit gives: \[ x \ln \left(1+\frac{2}{x}\right) \approx x \cdot \frac{2}{x} = 2. \] Now, we can formally take the limit: \[ \lim _{x \rightarrow+\infty} x \ln \left(1+\frac{2}{x}\right) = \lim _{x \rightarrow+\infty} 2 = 2. \] Thus, the final answer is: \[ \lim _{x \rightarrow+\infty} x \ln \left(1+\frac{2}{x}\right) = 2. \]