\( \lim _{x \rightarrow 0} (\frac{(1+x)^{\frac{1}{x}}-e}{x}) \)

We start with the limit \[ L = \lim_{x\to 0} \frac{(1+x)^{\frac{1}{x}} - e}{x}. \] **Step 1. Rewrite Using Exponential Function** Express \((1+x)^{\frac{1}{x}}\) as an exponential: \[ (1+x)^{\frac{1}{x}} = \exp\left(\frac{\ln(1+x)}{x}\right). \] **Step 2. Expand \(\ln(1+x)\) in a Taylor Series** We have the Taylor series expansion for \(\ln(1+x)\) around \(x=0\): \[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots. \] Dividing by \(x\) gives: \[ \frac{\ln(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3} - \cdots. \] **Step 3. Substitute Back into the Exponential** Substitute the expansion into the exponential expression: \[ (1+x)^{\frac{1}{x}} = \exp\left(1 - \frac{x}{2} + \frac{x^2}{3} - \cdots\right). \] This can be written as: \[ (1+x)^{\frac{1}{x}} = e \cdot \exp\left(- \frac{x}{2} + \frac{x^2}{3} - \cdots\right). \] **Step 4. Expand the Remaining Exponential** Expand \(\exp\left(-\frac{x}{2} + \frac{x^2}{3}\right)\) using its Taylor series: \[ \exp\left(-\frac{x}{2} + \frac{x^2}{3}\right) \approx 1 - \frac{x}{2} + \left(\frac{x^2}{3} + \frac{x^2}{8}\right) + \cdots, \] where the \(\frac{x^2}{8}\) term comes from \(\frac{1}{2!}\left(-\frac{x}{2}\right)^2\). Thus, \[ \exp\left(-\frac{x}{2} + \frac{x^2}{3}\right) \approx 1 - \frac{x}{2} + \frac{11x^2}{24} + \cdots. \] **Step 5. Substitute Back and Compute the Numerator** Now, \[ (1+x)^{\frac{1}{x}} \approx e \left(1 - \frac{x}{2} + \frac{11x^2}{24} + \cdots \right). \] Thus, the numerator becomes: \[ (1+x)^{\frac{1}{x}} - e \approx e\left(1 - \frac{x}{2} + \frac{11x^2}{24}\right) - e = e\left(-\frac{x}{2} + \frac{11x^2}{24}\right). \] **Step 6. Divide by \(x\) and Take the Limit** Divide by \(x\): \[ \frac{(1+x)^{\frac{1}{x}} - e}{x} \approx e\left(-\frac{1}{2} + \frac{11x}{24}\right). \] Taking the limit as \(x \to 0\): \[ \lim_{x\to 0} \frac{(1+x)^{\frac{1}{x}} - e}{x} = -\frac{e}{2}. \] Thus, the value of the limit is: \[ -\frac{e}{2}. \]