`lim_(xrarr0^(+))(e^(x)+x)^((1//x))`

Options

A. does not exist finitely
B. is 1
C. is `e^(2)`
D. is 2

Answer

To find the limit

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

we can start by rewriting the expression in a more manageable form.

### Step 1: Rewrite the limit using logarithms
Let \( y = (e^x + x)^{\frac{1}{x}} \). Taking the natural logarithm of both sides gives:

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

### Step 2: Find the limit of \(\ln y\)
Now we need to find:

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

### Step 3: Evaluate the limit using L'Hôpital's Rule
As \( x \to 0^+ \), both the numerator and denominator approach 0, which gives us a \( \frac{0}{0} \) form. We can apply L'Hôpital's Rule:

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

### Step 4: Differentiate the numerator and denominator
The derivative of the numerator is:

\[ \frac{d}{dx}[\ln(e^x + x)] = \frac{e^x + 1}{e^x + x}. \]

The derivative of the denominator is simply 1. Thus, we have:

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

### Step 5: Evaluate the limit
Now substituting \( x = 0 \):

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

### Step 6: Substitute back to find \( y \)
Since we found that

\[ \lim_{x \to 0^+} \ln y = 2, \]

we can exponentiate both sides to find \( y \):

\[ \lim_{x \to 0^+} y = e^2. \]

### Conclusion
Thus, the limit is:

\[ \lim_{x \to 0^+} (e^x + x)^{\frac{1}{x}} = e^2. \]

### Final Answer
The final answer is:

\[ e^2. \]