Slove `lim_(xto0)((1+x)^(1//x)-e)/x`

Answer

To solve the limit \( \lim_{x \to 0} \frac{(1+x)^{\frac{1}{x}} - e}{x} \), we will follow these steps:

### Step 1: Expand \( (1+x)^{\frac{1}{x}} \)
Using the known expansion for \( (1+x)^{\frac{1}{x}} \) as \( e \cdot \left( 1 - \frac{x}{2} + \frac{11x^2}{24} - \ldots \right) \), we can write:
\[ (1+x)^{\frac{1}{x}} \approx e \left( 1 - \frac{x}{2} + \frac{11x^2}{24} \right) \]

### Step 2: Substitute the expansion into the limit
Now, substituting this expansion into our limit expression:
\[ \lim_{x \to 0} \frac{(1+x)^{\frac{1}{x}} - e}{x} = \lim_{x \to 0} \frac{e \left( 1 - \frac{x}{2} + \frac{11x^2}{24} \right) - e}{x} \]
This simplifies to:
\[ \lim_{x \to 0} \frac{e - e + e \left( -\frac{x}{2} + \frac{11x^2}{24} \right)}{x} = \lim_{x \to 0} \frac{e \left( -\frac{x}{2} + \frac{11x^2}{24} \right)}{x} \]

### Step 3: Simplify the expression
Now, we can simplify the fraction:
\[ = \lim_{x \to 0} \left( -\frac{e}{2} + \frac{11ex}{24} \right) \]

### Step 4: Evaluate the limit
As \( x \to 0 \), the term \( \frac{11ex}{24} \) approaches 0. Therefore, we are left with:
\[ -\frac{e}{2} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{(1+x)^{\frac{1}{x}} - e}{x} = -\frac{e}{2} \]

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