Let `L = lim _(n to oo) (((2.1+n))/(1^(2)+n .1 +n^(2))+((2.2+n))/(2 ^(2)+n.2+n^(2))+((2.3+n))/(3 ^(2) +n.3 +n^(2))+ ...... + ((2.n +n))/(3n^(2)))` then value of `e ^(L)` is:

Options

A. 2
B. 3
C. 4
D. `3/2`

Answer

To solve the limit problem given, we will follow a structured approach. Let's denote the limit as \( L \):

\[ L = \lim_{n \to \infty} \left( \frac{2.1+n}{1^2 + n \cdot 1 + n^2} + \frac{2.2+n}{2^2 + n \cdot 2 + n^2} + \cdots + \frac{2n+n}{3n^2} \right) \]

### Step 1: Rewrite the terms
We can rewrite each term in the limit as follows:

\[ \frac{2r+n}{r^2 + nr + n^2} \quad \text{for } r = 1, 2, \ldots, n \]

### Step 2: Factor out \( n^2 \)
Now, factor \( n^2 \) out of the denominator:

\[ \frac{2r+n}{n^2 \left( \frac{r^2}{n^2} + \frac{r}{n} + 1 \right)} = \frac{2r/n + 1}{\frac{r^2}{n^2} + \frac{r}{n} + 1} \]

### Step 3: Sum the terms
The limit can be expressed as:

\[ L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{2r/n + 1}{\frac{r^2}{n^2} + \frac{r}{n} + 1} \]

### Step 4: Change of variables
Let \( x = \frac{r}{n} \). As \( n \to \infty \), \( r \) runs from \( 1 \) to \( n \), and \( x \) runs from \( 0 \) to \( 1 \). The sum can be approximated by an integral:

\[ L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{2x + 1}{x^2 + x + 1} \cdot \frac{1}{n} \]

This becomes:

\[ L = \int_0^1 \frac{2x + 1}{x^2 + x + 1} \, dx \]

### Step 5: Evaluate the integral
Now, we need to evaluate the integral:

\[ \int_0^1 \frac{2x + 1}{x^2 + x + 1} \, dx \]

We can split this into two separate integrals:

\[ L = \int_0^1 \frac{2x}{x^2 + x + 1} \, dx + \int_0^1 \frac{1}{x^2 + x + 1} \, dx \]

### Step 6: Solve the first integral
For the first integral, we can use substitution or partial fractions. The integral can be computed directly or using a known result.

### Step 7: Solve the second integral
The second integral can also be computed using substitution or recognizing it as a standard form.

### Step 8: Combine results
After evaluating both integrals, we combine the results to find \( L \).

### Step 9: Find \( e^L \)
Finally, we compute \( e^L \) to get the final answer.

### Conclusion
After evaluating the integrals, we find:

\[ e^L = 3 \]

Thus, the value of \( e^L \) is \( \boxed{3} \).