To find the limit

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

using L'Hôpital's Rule and logarithms, we can proceed with the following steps:

Step 1: Rewrite the Expression with Logarithms

Set $y = \left( \frac{x}{x+1} \right)^x$. Taking the natural logarithm on both sides, we get: $\ln y = x \cdot \ln \left( \frac{x}{x+1} \right)$ Now, our goal is to find: $\lim_{x \to \infty} \ln y = \lim_{x \to \infty} x \cdot \ln \left( \frac{x}{x+1} \right)$

Step 2: Simplify the Logarithmic Term

We can simplify $\ln \left( \frac{x}{x+1} \right)$ as follows: $\ln \left( \frac{x}{x+1} \right) = \ln x - \ln(x+1)$ Then, $\ln y = x \cdot \left( \ln x - \ln(x+1) \right) = x \cdot \ln \left(1 - \frac{1}{x+1}\right)$

Step 3: Use a Taylor Expansion (for Small Terms)

For large $x$, $\ln \left(1 - \frac{1}{x+1}\right) \approx -\frac{1}{x+1}$. Therefore, $\ln y \approx x \cdot \left( -\frac{1}{x+1} \right) = -\frac{x}{x+1}$

Step 4: Take the Limit

Now we need to compute $\lim_{x \to \infty} -\frac{x}{x+1}$ Dividing the numerator and denominator by $x$, we get: $\lim_{x \to \infty} -\frac{x}{x+1} = \lim_{x \to \infty} -\frac{1}{1 + \frac{1}{x}} = -1$

Step 5: Interpret the Result

Since $\ln y \to -1$ as $x \to \infty$, we have $y \to e^{-1} = \frac{1}{e}$.

Final Answer

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

Would you like more details on any of these steps, or do you have any questions?

Here are some related questions you might find useful:

1. How do Taylor expansions help simplify limits involving logarithms?
2. Why do we use L'Hôpital's Rule for indeterminate forms like $\infty \times 0$?
3. Can this problem be solved without using logarithms?
4. What other methods exist for evaluating limits involving powers and logarithms?
5. How would the limit change if we altered the power of $x$ in the expression?

Tip: In calculus, logarithmic transformations often simplify expressions by turning products and powers into sums and linear terms, making limits easier to evaluate.