1. **State the problem:** Evaluate the limit $$\lim_{x \to 0} (1 + \tan x)^{\cot x}$$. 2. **Recall the formula and rules:** This is a limit of the form $$\lim_{x \to 0} (1 + f(x))^{g(x)}$$ which often can be evaluated using the exponential and logarithm transformation: $$\lim_{x \to 0} (1 + f(x))^{g(x)} = e^{\lim_{x \to 0} g(x) \ln(1 + f(x))}$$. 3. **Apply the transformation:** Let $$f(x) = \tan x$$ and $$g(x) = \cot x = \frac{\cos x}{\sin x}$$. We write: $$\lim_{x \to 0} (1 + \tan x)^{\cot x} = e^{\lim_{x \to 0} \cot x \cdot \ln(1 + \tan x)}$$. 4. **Evaluate the inner limit:** As $$x \to 0$$, $$\tan x \approx x$$ and $$\ln(1 + \tan x) \approx \tan x$$ because $$\ln(1 + y) \approx y$$ for small $$y$$. So, $$\cot x \cdot \ln(1 + \tan x) \approx \frac{\cos x}{\sin x} \cdot \tan x = \frac{\cos x}{\sin x} \cdot \frac{\sin x}{\cos x} = 1$$. 5. **Confirm the limit:** More rigorously, $$\lim_{x \to 0} \cot x \cdot \ln(1 + \tan x) = \lim_{x \to 0} \frac{\cos x}{\sin x} \cdot \ln(1 + \tan x)$$ Using the approximations: $$\cot x \approx \frac{1}{x}$$ $$\ln(1 + \tan x) \approx x$$ So, $$\lim_{x \to 0} \cot x \cdot \ln(1 + \tan x) = \lim_{x \to 0} \frac{1}{x} \cdot x = 1$$. 6. **Final answer:** $$\lim_{x \to 0} (1 + \tan x)^{\cot x} = e^{1} = e$$.