The given mathematical expression appears to define the limit of a sequence $S_n$ as $n \to \infty$. Specifically:

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

Let’s solve this limit step-by-step:

Step 1: Analyze the fraction

The sequence $S_n = \frac{2n}{n+3}$ is a rational function where the degree of the numerator and denominator is the same (both have degree 1 in $n$).

Step 2: Divide numerator and denominator by the highest degree term ($n$):

$\frac{2n}{n+3} = \frac{2n}{n \cdot \left(1 + \frac{3}{n}\right)} = \frac{2}{1 + \frac{3}{n}}.$

Step 3: Take the limit as $n \to \infty$:

As $n \to \infty$, the term $\frac{3}{n} \to 0$. Therefore, the fraction simplifies to: $\lim_{n \to \infty} \frac{2}{1 + \frac{3}{n}} = \frac{2}{1 + 0} = 2.$

Final Answer:

$L = 2.$

The sequence $S_n$ converges to $L = 2$ as $n \to \infty$.

Would you like a detailed explanation of any step? Here are some related questions:

1. What happens to terms like $\frac{3}{n}$ as $n \to \infty$?
2. How do limits of rational functions behave when the degrees of numerator and denominator are equal?
3. Could we use L’Hôpital’s Rule to confirm this limit?
4. What would happen if the numerator had a higher degree than the denominator?
5. How can this method apply to other similar sequences?

Tip: For rational functions, always simplify by dividing numerator and denominator by the highest degree term in $n$ to find the limit.