We are tasked with finding:

$\lim_{x \to +\infty} \left( \frac{x+1}{x+2} \right)^x$

Step 1: Simplify the base expression

We can rewrite the base expression:

$\frac{x+1}{x+2} = 1 - \frac{1}{x+2}$

Thus, the limit becomes:

$\lim_{x \to +\infty} \left( 1 - \frac{1}{x+2} \right)^x$

Step 2: Apply logarithms

To deal with the exponent $x$, let's take the natural logarithm of the expression. Let:

$L = \lim_{x \to +\infty} \left( 1 - \frac{1}{x+2} \right)^x$

Now take the natural logarithm of both sides:

$\ln(L) = \lim_{x \to +\infty} x \ln\left( 1 - \frac{1}{x+2} \right)$

Step 3: Use the approximation for $\ln(1 - y)$

For small $y$, we know that $\ln(1 - y) \approx -y$. Here, $y = \frac{1}{x+2}$, which becomes very small as $x \to +\infty$. Therefore, we approximate:

$\ln\left( 1 - \frac{1}{x+2} \right) \approx -\frac{1}{x+2}$

Thus, the logarithmic limit becomes:

$\ln(L) = \lim_{x \to +\infty} x \left( -\frac{1}{x+2} \right)$

Step 4: Simplify the limit

As $x \to +\infty$, $x+2 \approx x$, so we have:

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

Step 5: Exponentiate to find $L$

Now, exponentiate both sides to find $L$:

$L = e^{-1} = \frac{1}{e}$

Conclusion:

The limit is:

$\lim_{x \to +\infty} \left( \frac{x+1}{x+2} \right)^x = \frac{1}{e}$

Would you like more details or clarifications on any step?

Here are 5 related questions:

1. How can logarithms simplify limits involving exponents?
2. Why does the approximation $\ln(1 - y) \approx -y$ hold for small $y$?
3. What are other methods for evaluating limits of the form $\lim_{x \to +\infty} f(x)^x$?
4. How does the behavior of functions at infinity affect their limits?
5. How can the binomial expansion help in evaluating limits?

Tip: When dealing with limits that involve expressions raised to infinity, logarithms can often simplify the calculations considerably.