The image contains the following mathematical limit problem:

$\lim_{{n \to \infty}} \frac{3n^2 + 2n + 1}{n^2 + 1} = 3$

To solve this limit, let's simplify the expression:

Step 1: Divide by the highest power of $n^2$

Both the numerator and the denominator have terms involving $n^2$, so we divide each term by $n^2$:

$\frac{3n^2 + 2n + 1}{n^2 + 1} = \frac{3 + \frac{2}{n} + \frac{1}{n^2}}{1 + \frac{1}{n^2}}$

Step 2: Evaluate the limit as $n \to \infty$

As $n$ approaches infinity:

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

So the expression simplifies to:

$\frac{3 + 0 + 0}{1 + 0} = \frac{3}{1} = 3$

Thus, the limit is:

$\lim_{{n \to \infty}} \frac{3n^2 + 2n + 1}{n^2 + 1} = 3$

This confirms the given equation is correct.

Would you like more details or have any questions?

Related Questions:

1. How would the result change if the highest degree terms were different?
2. Can this limit be computed using L'Hôpital's Rule instead?
3. How would you approach this limit if the numerator was of lower degree than the denominator?
4. What is the significance of dividing by the highest power of $n$ in limits?
5. How can we generalize this method for other rational functions?

Tip:

When dealing with limits of rational functions, always focus on the highest degree terms as they dominate the behavior for large values of $n$.