The value of `lim_(xrarr0) (log_e(1+x)-x)/(x{(1+x)^(1//x)-e})` equal to

Options

A. `e^e`
B. `e`
C. `1//e`
D. `1`

Explanation

To find the value of the limit

\[ \lim_{x \to 0} \frac{\log(1+x) - x}{x \left( (1+x)^{\frac{1}{x}} - e \right)}, \]

we will use the Taylor series expansion for \(\log(1+x)\) and analyze the expression step by step.

### Step 1: Expand \(\log(1+x)\)

The Taylor series expansion for \(\log(1+x)\) around \(x=0\) is:

\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

### Step 2: Substitute the expansion into the limit

Substituting this expansion into the limit gives:

\[ \log(1+x) - x = \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \right) - x = -\frac{x^2}{2} + \frac{x^3}{3} - \ldots \]

Thus, the numerator becomes:

\[ -\frac{x^2}{2} + O(x^3) \]

### Step 3: Analyze the denominator

Next, we need to analyze the denominator \(x \left( (1+x)^{\frac{1}{x}} - e \right)\).

Using the limit definition, we know that:

\[ \lim_{x \to 0} (1+x)^{\frac{1}{x}} = e. \]

To find the expansion of \((1+x)^{\frac{1}{x}}\) as \(x \to 0\), we can rewrite it as:

\[ (1+x)^{\frac{1}{x}} = e^{\frac{\log(1+x)}{x}}. \]

Using the expansion for \(\log(1+x)\):

\[ \frac{\log(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3} - \ldots \]

Thus, we have:

\[ (1+x)^{\frac{1}{x}} \approx e^{1 - \frac{x}{2} + O(x^2)} = e \cdot e^{-\frac{x}{2} + O(x^2)}. \]

Using the expansion \(e^y \approx 1 + y\) for small \(y\):

\[ e^{-\frac{x}{2} + O(x^2)} \approx 1 - \frac{x}{2} + O(x^2). \]

Therefore:

\[ (1+x)^{\frac{1}{x}} - e \approx e \left( -\frac{x}{2} + O(x^2) \right). \]

### Step 4: Substitute back into the limit

Now substituting this back into our limit, we have:

\[ x \left( (1+x)^{\frac{1}{x}} - e \right) \approx x \left( -\frac{ex}{2} + O(x^2) \right) = -\frac{ex^2}{2} + O(x^3). \]

### Step 5: Combine the results

Now substituting both the numerator and denominator into the limit:

\[ \lim_{x \to 0} \frac{-\frac{x^2}{2} + O(x^3)}{-\frac{ex^2}{2} + O(x^3)}. \]

As \(x \to 0\), the \(O(x^3)\) terms become negligible, and we can simplify:

\[ \lim_{x \to 0} \frac{-\frac{x^2}{2}}{-\frac{ex^2}{2}} = \lim_{x \to 0} \frac{1}{e} = \frac{1}{e}. \]

### Final Answer

Thus, the value of the limit is:

\[ \frac{1}{e}. \]