The limit `L=lim_(nrarroo)Sigma_(r=4)^(n-4)(n)/(n^(2)+r^(2))` satisfies the relation

Options

A. `e^(L) gt e`
B. `e^(L) gt 3`
C. `e^(tanL) lt 2e`
D. `(pi)/(L) lt 1`

Answer

To solve the limit \( L = \lim_{n \to \infty} \sum_{r=4}^{n-4} \frac{n}{n^2 + r^2} \), we will follow these steps:

### Step 1: Rewrite the Summation
We start with the expression for \( L \):
\[ L = \lim_{n \to \infty} \sum_{r=4}^{n-4} \frac{n}{n^2 + r^2} \]
We can factor out \( n^2 \) from the denominator:
\[ L = \lim_{n \to \infty} \sum_{r=4}^{n-4} \frac{n}{n^2(1 + \frac{r^2}{n^2})} = \lim_{n \to \infty} \sum_{r=4}^{n-4} \frac{1}{n(1 + \frac{r^2}{n^2})} \]

### Step 2: Change of Variables
We can use the substitution \( x = \frac{r}{n} \), which implies \( r = nx \). The limits of summation change accordingly:
- When \( r = 4 \), \( x = \frac{4}{n} \)
- When \( r = n-4 \), \( x = 1 - \frac{4}{n} \)

Thus, we can rewrite the sum as:
\[ L = \lim_{n \to \infty} \sum_{r=4}^{n-4} \frac{1}{n(1 + \frac{(nx)^2}{n^2})} = \lim_{n \to \infty} \sum_{r=4}^{n-4} \frac{1}{n(1 + x^2)} \]

### Step 3: Convert the Sum to an Integral
As \( n \to \infty \), the sum approaches the integral:
\[ L = \int_{0}^{1} \frac{1}{1 + x^2} \, dx \]

### Step 4: Evaluate the Integral
The integral \( \int \frac{1}{1 + x^2} \, dx \) is known to be \( \tan^{-1}(x) \). Evaluating from 0 to 1 gives:
\[ L = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### Step 5: Analyze the Options
Now that we have \( L = \frac{\pi}{4} \), we can analyze the options given in the problem statement:

1. \( e^{\frac{\pi}{4}} \) is approximately \( 2.19 \).
2. \( e^L > 3 \) is false since \( e^{\frac{\pi}{4}} < 3 \).
3. \( e^{10L} < 2e \) is true since \( e^{10 \cdot \frac{\pi}{4}} < 2e \).
4. \( \frac{\pi}{L} < 1 \) is false since \( \frac{\pi}{\frac{\pi}{4}} = 4 \).

Thus, the only correct relation is option 3.

### Final Answer
The limit \( L = \frac{\pi}{4} \) satisfies the relation \( e^{10L} < 2e \).

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