1. **Stating the problem:** Calculate the limit $$\lim_{x \to -\infty} \left(\frac{x + 4}{x + 2}\right)^x.$$\n\n2. **Recall the formula and rules:** For limits of the form $$\lim_{x \to \pm\infty} \left(1 + \frac{a}{x}\right)^x = e^a,$$ we try to rewrite the expression in a similar form.\n\n3. **Rewrite the expression:**\n\n$$\frac{x + 4}{x + 2} = \frac{x + 2 + 2}{x + 2} = 1 + \frac{2}{x + 2}.$$\n\nSo the limit becomes $$\lim_{x \to -\infty} \left(1 + \frac{2}{x + 2}\right)^x.$$\n\n4. **Change variable for clarity:** Let $$t = x + 2,$$ so as $$x \to -\infty,$$ also $$t \to -\infty.$$ The expression is now $$\left(1 + \frac{2}{t}\right)^{t - 2} = \left(1 + \frac{2}{t}\right)^t \cdot \left(1 + \frac{2}{t}\right)^{-2}.$$\n\n5. **Evaluate the limit:**\n\n- First, $$\lim_{t \to -\infty} \left(1 + \frac{2}{t}\right)^t = e^2$$ because the base tends to 1 and the exponent tends to $$-\infty$$, but since $$t$$ is negative, the limit is $$e^2$$.\n- Second, $$\lim_{t \to -\infty} \left(1 + \frac{2}{t}\right)^{-2} = 1^{-2} = 1.$$\n\n6. **Combine the results:**\n\n$$\lim_{x \to -\infty} \left(\frac{x + 4}{x + 2}\right)^x = e^2 \cdot 1 = e^2.$$\n\n**Final answer:** $$e^2.$$