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 $0^+$.

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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.