To solve the limit

$\lim_{{x \to \infty}} \frac{x+1}{x+2}$

we'll analyze the expression by dividing the numerator and the denominator by $x$:

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

As $x \to \infty$, the terms $\frac{1}{x}$ and $\frac{2}{x}$ approach zero. Therefore, the expression simplifies to:

$\lim_{{x \to \infty}} \frac{1 + \frac{1}{x}}{1 + \frac{2}{x}} = \frac{1 + 0}{1 + 0} = \frac{1}{1} = 1.$

So,

$\lim_{{x \to \infty}} \frac{x+1}{x+2} = 1.$

Would you like more details or have any questions?

Here are some related questions that might interest you:

1. How do you find limits involving rational functions in general?
2. What is the behavior of functions like $\frac{x + 1}{x + n}$ as $x \to \infty$ for different values of $n$?
3. How can L'Hôpital's Rule be applied to similar limits?
4. What other techniques are useful for limits as $x \to \infty$ in more complex expressions?
5. How can this limit be interpreted in terms of horizontal asymptotes?

Tip: For rational functions, dividing terms by the highest power of $x$ can simplify evaluating limits as $x \to \infty$.