The sequence given in the image is:

$u_n = \frac{2n + 1}{n + 325}$

To find the limit of this sequence as $n$ approaches infinity ($n \to \infty$), let's analyze it step by step.

Step 1: Divide Numerator and Denominator by $n$

$u_n = \frac{2n + 1}{n + 325} = \frac{2n/n + 1/n}{n/n + 325/n} = \frac{2 + \frac{1}{n}}{1 + \frac{325}{n}}$

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

As $n$ becomes very large, the terms $\frac{1}{n}$ and $\frac{325}{n}$ approach 0. So, the expression simplifies to: $\lim_{n \to \infty} u_n = \frac{2 + 0}{1 + 0} = \frac{2}{1} = 2$

Conclusion:

The limit of the sequence $u_n$ as $n$ approaches infinity is $2$.

Do you need further details or have any other questions?

Here are five related questions to expand on this topic:

1. How does the limit of a sequence relate to the concept of convergence?
2. Can you apply the same method to find the limit of a different sequence, like $\frac{3n^2 + n}{n^2 + 5n + 1}$?
3. What happens if the sequence includes higher powers of $n$ in the numerator or denominator?
4. How does the presence of constants, like 325 in the denominator, affect the limit?
5. What is the significance of determining the limit of a sequence in mathematical analysis?

Tip: When finding the limit of a sequence, dividing both the numerator and the denominator by the highest power of $n$ often simplifies the expression and makes it easier to determine the limit.