We are tasked with finding the limit of the following expression as $n \to \infty$:

$\lim_{n \to \infty} \frac{n+2}{n^2 + 1}.$

Step-by-Step Solution:

1. Expression Review: The expression is $\frac{n+2}{n^2 + 1}$. As $n$ grows large, both the numerator and the denominator increase, but at different rates. The numerator $n+2$ grows linearly (like $n$), while the denominator $n^2+1$ grows quadratically (like $n^2$).

2. Divide by the Highest Power of $n$: To make the behavior at infinity clearer, divide both the numerator and denominator by $n^2$ (the highest power of $n$ in the denominator):

   $\frac{n+2}{n^2+1} = \frac{\frac{n+2}{n^2}}{\frac{n^2+1}{n^2}} = \frac{\frac{n}{n^2} + \frac{2}{n^2}}{1 + \frac{1}{n^2}}.$

3. Simplify the Expression: Simplifying each term, we get:

   $= \frac{\frac{1}{n} + \frac{2}{n^2}}{1 + \frac{1}{n^2}}.$

4. Take the Limit as $n \to \infty$: Now, take the limit as $n \to \infty$:

   • $\frac{1}{n} \to 0$,
   • $\frac{2}{n^2} \to 0$,
   • $\frac{1}{n^2} \to 0$.

   So the expression becomes:

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

Conclusion:

Thus, we have shown that:

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

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

Here are 5 related questions to deepen your understanding:

1. How would the limit change if the numerator were $n^2+2$?
2. What happens to the limit if the denominator were $n+1$ instead of $n^2+1$?
3. Can we apply L'Hopital's Rule to this problem? Why or why not?
4. What is the limit of $\frac{n^3 + 2}{n^4 + 1}$ as $n \to \infty$?
5. How would you compute the limit of a more complicated rational function?

Tip: Dividing both the numerator and denominator by the highest power of $n$ helps simplify the behavior of limits at infinity.