To evaluate the limit

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

using L'Hôpital's Rule, let's follow these steps:

Step 1: Rewrite the Expression in a More Convenient Form

This expression $\left(1 + x\right)^{\frac{1}{x}}$ has an indeterminate form of $1^{\infty}$ as $x \to 0^+$. To handle this, we can take the natural logarithm of the expression, which allows us to transform it into a limit involving products.

Define $y = \left(1 + x\right)^{\frac{1}{x}}.$

Taking the natural logarithm of both sides, we get: $\ln y = \frac{1}{x} \ln(1 + x).$

Now, our goal is to find the limit of $\ln y$ as $x \to 0^+$, which becomes: $\lim_{x \to 0^+} \frac{\ln(1 + x)}{x}.$

Step 2: Apply L'Hôpital's Rule

The limit $\lim_{x \to 0^+} \frac{\ln(1 + x)}{x}$ is of the form $\frac{0}{0}$, so we can apply L'Hôpital's Rule. Differentiate the numerator and the denominator with respect to $x$:

1. The derivative of the numerator, $\ln(1 + x)$, is $\frac{1}{1 + x}$.
2. The derivative of the denominator, $x$, is $1$.

Thus, we get: $\lim_{x \to 0^+} \frac{\ln(1 + x)}{x} = \lim_{x \to 0^+} \frac{\frac{1}{1 + x}}{1} = \lim_{x \to 0^+} \frac{1}{1 + x}.$

As $x \to 0^+$, $\frac{1}{1 + x} \to 1$. So, we have: $\lim_{x \to 0^+} \frac{\ln(1 + x)}{x} = 1.$

Step 3: Exponentiate to Find the Original Limit

Since $\ln y \to 1$ as $x \to 0^+$, we have: $y \to e^1 = e.$

Therefore: $\lim_{x \to 0^+} \left(1 + x\right)^{\frac{1}{x}} = e.$

Final Answer:

$\lim_{x \to 0^+} \left(1 + x\right)^{\frac{1}{x}} = e.$

Would you like a more detailed breakdown or have any questions?

Here are some related questions to expand on this topic:

1. How does L'Hôpital's Rule work with other indeterminate forms like $0/0$ or $\infty/\infty$?
2. Can we apply L'Hôpital's Rule to limits involving exponential functions?
3. How can we evaluate limits without L'Hôpital's Rule, for example, using series expansion?
4. What are other methods for dealing with indeterminate forms like $1^{\infty}$ or $0^0$?
5. How does this limit relate to the definition of the exponential function $e$?

Tip: When dealing with indeterminate forms involving exponents, taking the logarithm often simplifies the problem by transforming multiplication into addition.