The provided mathematical expression seems to calculate the limit of a sequence:

$L = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{2n}{n+3}.$

To compute this limit:

Step-by-Step Solution:

1. Divide numerator and denominator by $n$ (the highest degree of $n$ in the denominator): $\frac{2n}{n+3} = \frac{2n}{n(1 + \frac{3}{n})}.$

2. Simplify the expression: $\frac{2n}{n(1 + \frac{3}{n})} = \frac{2}{1 + \frac{3}{n}}.$

3. Take the limit as $n \to \infty$: As $n \to \infty$, $\frac{3}{n} \to 0$. Thus: $\frac{2}{1 + \frac{3}{n}} \to \frac{2}{1 + 0} = 2.$

Final Result:

$L = 2.$

Let me know if you want additional details or explanations!

Expanding on This:

1. What is the general rule for finding limits of rational functions as $n \to \infty$?
2. How would the limit change if the numerator's degree was higher than the denominator's?
3. How does this limit relate to convergence in sequences?
4. Could this problem be extended to integrals involving rational functions?
5. How would you verify this limit numerically?

Tip: Always simplify expressions by factoring out the dominant term in the denominator when $n \to \infty$.