To solve the limit

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

using logarithmic and L'Hôpital's Rule, let's break it down into steps.

1. Rewrite the Expression Using Logarithms: We set $y = \left(\frac{x}{x+1}\right)^x$ and take the natural logarithm of both sides to simplify the power:

   $\ln y = x \ln \left(\frac{x}{x+1}\right)$

   Now, we need to evaluate

   $\lim_{x \to \infty} \ln y = \lim_{x \to \infty} x \ln \left(\frac{x}{x+1}\right)$

2. Simplify the Logarithmic Term: Rewrite $\ln \left(\frac{x}{x+1}\right)$:

   $\ln \left(\frac{x}{x+1}\right) = \ln x - \ln(x+1) = \ln \left(1 - \frac{1}{x+1}\right)$

   As $x \to \infty$, $\ln \left(\frac{x}{x+1}\right)$ approaches $\ln(1 - 0) = \ln(1) = 0$. However, we need to handle the expression in a way that allows us to evaluate the limit correctly.

3. Rewrite as a Product Suitable for L'Hôpital's Rule: We have:

   $\lim_{x \to \infty} x \ln \left(\frac{x}{x+1}\right) = \lim_{x \to \infty} x \ln \left(1 - \frac{1}{x+1}\right)$

   Using the approximation $\ln(1 - u) \approx -u$ when $u$ is small, we get:

   $\ln \left(1 - \frac{1}{x+1}\right) \approx -\frac{1}{x+1}$

   Thus,

   $\lim_{x \to \infty} x \ln \left(\frac{x}{x+1}\right) \approx \lim_{x \to \infty} -\frac{x}{x+1} = -1$

4. Interpret the Result: Since $\lim_{x \to \infty} \ln y = -1$, we have:

   $\ln y = -1 \implies y = e^{-1} = \frac{1}{e}$

Therefore,

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

The final answer is:

$\frac{1}{e}$

Would you like further clarification on any of the steps?

Here are some additional questions to expand on this problem:

1. What happens if the exponent grows faster or slower than $x$?
2. How can we apply similar techniques to limits with logarithmic or exponential expressions?
3. What is the significance of the approximation $\ln(1 - u) \approx -u$ in calculus?
4. How does this technique apply to expressions involving fractions raised to powers in different contexts?
5. Can we derive alternative solutions using other forms of L'Hôpital’s Rule?

Tip: Using logarithmic transformations in limit problems involving exponentiation simplifies handling indeterminate forms.