please give a full explanation: \lim _{x\to 0+}\left(e^{\frac{lnxe}{x}}\right)

To evaluate the limit \(\lim _{x\to 0^+}\left(e^{\frac{\ln(xe)}{x}}\right)\), let's follow these steps:

### Step 1: Simplify the Expression Inside the Exponential
First, consider the expression inside the exponential function:

\[
\frac{\ln(xe)}{x}
\]

Notice that \(\ln(xe)\) can be expanded using the logarithm properties:

\[
\ln(xe) = \ln(x) + \ln(e)
\]

Since \(\ln(e) = 1\), we have:

\[
\ln(xe) = \ln(x) + 1
\]

Thus, the expression inside the limit becomes:

\[
\frac{\ln(x) + 1}{x}
\]

So the limit can be rewritten as:

\[
\lim _{x\to 0^+}\left(e^{\frac{\ln(x) + 1}{x}}\right)
\]

### Step 2: Analyze the Limit of the Exponent
Next, let's analyze the limit of the exponent \(\frac{\ln(x) + 1}{x}\) as \(x \to 0^+\):

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

To understand this limit, break it into two parts:

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

#### Part 1: \(\lim_{x \to 0^+} \frac{\ln(x)}{x}\)
As \(x \to 0^+\), \(\ln(x)\) approaches \(-\infty\), and \(\frac{\ln(x)}{x}\) involves a \( -\infty \) term divided by a small positive \( x \). This term tends to \(-\infty\):

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

#### Part 2: \(\lim_{x \to 0^+} \frac{1}{x}\)
As \(x \to 0^+\), \(\frac{1}{x}\) grows without bound, meaning it tends to \(+\infty\):

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

### Step 3: Combine the Results of Both Limits
Now, let's combine the two parts:

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

This expression is an indeterminate form. However, considering that the \(\frac{1}{x}\) term dominates (since \(\ln(x)\) grows much more slowly compared to \(1/x\) as \(x \to 0^+\)), the sum tends to \(+\infty\).

### Step 4: Evaluate the Original Limit
Now that we know the exponent \(\frac{\ln(x) + 1}{x}\) tends to \(+\infty\) as \(x \to 0^+\), we substitute this into the original limit:

\[
\lim _{x\to 0^+}\left(e^{\frac{\ln(xe)}{x}}\right) = e^{+\infty}
\]

Since \(e^{+\infty}\) grows without bound, the limit is:

\[
\lim _{x\to 0^+}\left(e^{\frac{\ln(xe)}{x}}\right) = \boxed{\infty}
\]

### Conclusion
Thus, the limit evaluates to \(\infty\) as \(x\) approaches \(0^+\).

---

Would you like to delve deeper into any part of this explanation? Here are some related questions to consider:

1. How do you evaluate \(\lim _{x\to 0^+} \frac{\ln(x)}{x}\)?
2. What are the different types of indeterminate forms in limits?
3. How can L'Hôpital's Rule be applied to the limit \(\lim _{x\to 0^+} \frac{\ln(x)}{x}\)?
4. What happens if you change the limit to \(\lim _{x\to 0^-}\)?
5. How do logarithms behave near 0 and near infinity?
6. What is the limit \(\lim _{x\to \infty} \frac{\ln(x)}{x}\)?
7. How does the behavior of the function change if the base of the exponential is different?
8. Can you analyze \(\lim _{x\to 0^+} e^{\frac{\ln(x)}{x}}\)?

**Tip:** When dealing with limits, especially involving logarithms and exponential functions, always consider if the expression is an indeterminate form before proceeding to simplify or apply limit laws.

Understand how to evaluate the limit \(\lim _{x\to 0^+}\left(e^{\frac{\ln(xe)}{x}}\right)\) step-by-step. Learn about limits involving exponential and logarithmic functions, indeterminate forms, and the behavior of logarithms near zero. Suitable for advanced high school students and anyone interested in calculus and mathematical analysis.

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