The given expression is:

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

To solve this, we need to analyze the limit carefully. Notice that as \( x \to 0 \), the term \( \left(1 + \frac{1}{x}\right) \) becomes problematic, since \( \frac{1}{x} \) approaches infinity when \( x \to 0^+ \) (from the positive side) and negative infinity when \( x \to 0^- \) (from the negative side).

Let's consider the behavior of the function separately as \( x \to 0^+ \) and \( x \to 0^- \):

1. **As \( x \to 0^+ \)**: In this case, \( \frac{1}{x} \) becomes very large, so the expression \( \left(1 + \frac{1}{x}\right)^x \) approaches 0.

2. **As \( x \to 0^- \)**: For negative values of \( x \), \( \frac{1}{x} \) becomes a large negative number, so \( \left(1 + \frac{1}{x}\right)^x \) becomes undefined because the base inside the parentheses becomes negative, leading to complex values.

Thus, this limit does not exist as a real number because the expression behaves differently as \( x \to 0^+ \) and \( x \to 0^- \), and on the negative side, it can even lead to complex values.

Would you like more details or have any questions?

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Here are five related questions to expand your understanding:

1. What happens if we replace the \( \frac{1}{x} \) term with a different fraction?
2. How can we handle limits approaching from only one side, like \( x \to 0^+ \)?
3. What types of limits lead to complex results in general?
4. How would the behavior change if the exponent were something other than \( x \)?
5. How does the function behave at other critical points like \( x \to \infty \)?

**Tip:** For tricky limits involving powers and fractions, consider using logarithms to simplify the expression.

Explore the limit of the function (1 + 1/x)^x as x approaches 0. Understand how the function behaves from both positive and negative sides and why the limit does not exist as a real number. This concept is crucial in calculus and precalculus.

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