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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 \( x \to \infty \).

Evaluate the limit \( \lim_{{x \to \infty}} \frac{x+1}{x+2} \).

Learn how to evaluate the limit of the rational function \( \lim_{{x \to \infty}} \frac{x+1}{x+2} \) as \( x \) approaches infinity. This example covers asymptotic behavior and is suitable for students in Grades 11-12.

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