limit of (x+e^x)^(1/x) when x approaches 0

To find the limit of \((x + e^x)^{1/x}\) as \(x\) approaches 0, we can use some algebraic manipulation and calculus techniques. Let's proceed with the solution:

### Problem:
\[
\lim_{x \to 0} (x + e^x)^{1/x}
\]

### Step-by-Step Solution:

1. **Rewrite the Expression:**

   To make this easier to handle, let's take the natural logarithm of the expression:

   \[
   y = (x + e^x)^{1/x}
   \]

   Taking the natural logarithm on both sides:

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

   Now, we want to find:

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

   If we can find this limit, we can then exponentiate the result to find \(\lim_{x \to 0} y\).

2. **Apply L'Hôpital's Rule:**

   The limit \(\lim_{x \to 0} \frac{\ln(x + e^x)}{x}\) is of the indeterminate form \(\frac{0}{0}\). We can apply L'Hôpital's rule, which states that if the limit is indeterminate, then:

   \[
   \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)}
   \]

   Here, \(f(x) = \ln(x + e^x)\) and \(g(x) = x\).

   Find the derivatives:

   - \(f'(x) = \frac{1}{x + e^x} \cdot (1 + e^x) = \frac{1 + e^x}{x + e^x}\)
   - \(g'(x) = 1\)

   Applying L'Hôpital's rule:

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

3. **Evaluate the Limit:**

   As \(x\) approaches \(0\), \(e^x\) approaches \(1\). So:

   - \(1 + e^x \to 2\)
   - \(x + e^x \to 1\)

   Therefore, the limit becomes:

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

   Hence, we have:

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

4. **Exponentiate to Find the Final Answer:**

   Since \(\ln y\) approaches 2 as \(x\) approaches 0:

   \[
   y = e^{\ln y} \Rightarrow \lim_{x \to 0} y = e^2
   \]

### Conclusion:

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

Would you like more details or have any questions? Here are some related questions to expand your understanding:

1. What is the limit of \((1 + x)^{1/x}\) as \(x\) approaches 0?
2. How does L'Hôpital's Rule apply to different indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)?
3. Can we find the limit of \((x + e^x)^{1/x}\) using Taylor series expansion instead?
4. What is the behavior of the function \((x + e^x)^{1/x}\) as \(x\) approaches infinity?
5. How can you determine if a limit will approach \(e\) using the definition of \(e\) as a limit?

**Tip:** Taking the logarithm of an expression often simplifies the process of finding limits, especially when dealing with exponents and indeterminate forms.

Explore the limit of (x + e^x)^(1/x) as x approaches 0 with a detailed step-by-step solution using algebraic manipulation and calculus techniques. Understand the application of L'Hôpital's Rule in evaluating limits involving exponential and logarithmic functions.

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