1. **State the problem:** We want to evaluate the limit $$\lim_{x \to 0} \left((1+x)^{\frac{1}{x}} - e^x\right).$$ 2. **Recall the formulas and rules:** - The expression $(1+x)^{\frac{1}{x}}$ is known to approach $e$ as $x \to 0$. - The exponential function $e^x$ also approaches $1$ as $x \to 0$. - Since direct substitution gives $e - 1$, but we have a difference of two expressions both approaching $e$, we need to analyze the limit carefully. - L’Hôpital’s Rule applies to limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. 3. **Rewrite the limit to apply L’Hôpital’s Rule:** We rewrite the limit as $$ \lim_{x \to 0} \left((1+x)^{\frac{1}{x}} - e^x\right) = \lim_{x \to 0} \frac{(1+x)^{\frac{1}{x}} - e^x}{1}. $$ This is not a quotient, so we consider the expansions or transform it into a quotient form. 4. **Use series expansions:** - Expand $(1+x)^{\frac{1}{x}}$ using the exponential and logarithm: $$ (1+x)^{\frac{1}{x}} = e^{\frac{1}{x} \ln(1+x)}. $$ - Use the expansion $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$. 5. **Calculate the exponent:** $$ \frac{1}{x} \ln(1+x) = \frac{1}{x} \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \right) = 1 - \frac{x}{2} + \frac{x^2}{3} - \cdots $$ 6. **Rewrite $(1+x)^{1/x}$:** $$ (1+x)^{\frac{1}{x}} = e^{1 - \frac{x}{2} + \frac{x^2}{3} - \cdots} = e \cdot e^{-\frac{x}{2} + \frac{x^2}{3} - \cdots}. $$ 7. **Expand $e^{-\frac{x}{2} + \frac{x^2}{3} - \cdots}$:** Using $e^y \approx 1 + y + \frac{y^2}{2} + \cdots$ for small $y$, $$ e^{-\frac{x}{2} + \frac{x^2}{3}} \approx 1 - \frac{x}{2} + \frac{x^2}{3} + \frac{1}{2} \left(-\frac{x}{2}\right)^2 = 1 - \frac{x}{2} + \frac{x^2}{3} + \frac{x^2}{8} = 1 - \frac{x}{2} + \frac{11x^2}{24}. $$ 8. **So, $(1+x)^{1/x}$ approximates:** $$ (1+x)^{\frac{1}{x}} \approx e \left(1 - \frac{x}{2} + \frac{11x^2}{24}\right) = e - \frac{e x}{2} + \frac{11 e x^2}{24}. $$ 9. **Expand $e^x$:** $$ e^x = 1 + x + \frac{x^2}{2} + \cdots. $$ 10. **Calculate the difference:** $$ (1+x)^{\frac{1}{x}} - e^x \approx \left(e - \frac{e x}{2} + \frac{11 e x^2}{24}\right) - \left(1 + x + \frac{x^2}{2}\right). $$ 11. **Evaluate the limit as $x \to 0$:** - Constant terms: $e - 1$. - Linear terms: $-\frac{e x}{2} - x = x\left(-\frac{e}{2} - 1\right)$. - Quadratic terms: $\frac{11 e x^2}{24} - \frac{x^2}{2} = x^2 \left(\frac{11 e}{24} - \frac{1}{2}\right)$. 12. **Since the limit is as $x \to 0$, the dominant term is the constant:** $$ \lim_{x \to 0} \left((1+x)^{\frac{1}{x}} - e^x\right) = e - 1. $$ **Final answer:** $$ \boxed{e - 1}. $$