1. **State the problem:** Evaluate the limit $$\lim_{x \to \infty} \left(1 + x\right)^{\frac{2}{x}}.$$\n\n2. **Recall the formula and important rules:** The expression resembles the form $$\lim_{x \to \infty} \left(1 + \frac{1}{n}\right)^n = e,$$ but here the base and exponent are different. We need to analyze the behavior as $$x \to \infty$$ carefully.\n\n3. **Rewrite the expression:**\n$$\left(1 + x\right)^{\frac{2}{x}} = e^{\ln\left(\left(1 + x\right)^{\frac{2}{x}}\right)} = e^{\frac{2}{x} \ln(1 + x)}.$$\n\n4. **Evaluate the exponent limit:**\nWe focus on $$\lim_{x \to \infty} \frac{2}{x} \ln(1 + x).$$\nAs $$x \to \infty,$$ $$\ln(1 + x) \sim \ln x,$$ so the limit becomes $$\lim_{x \to \infty} \frac{2 \ln x}{x}.$$\n\n5. **Apply limit:**\nSince $$\ln x$$ grows slower than any power of $$x,$$ $$\lim_{x \to \infty} \frac{\ln x}{x} = 0.$$\nTherefore, $$\lim_{x \to \infty} \frac{2 \ln x}{x} = 0.$$\n\n6. **Conclude the limit:**\n$$\lim_{x \to \infty} \left(1 + x\right)^{\frac{2}{x}} = e^0 = 1.$$\n\n**Final answer:** $$1.$$