Let `L= lim_(nrarr infty) int_(a)^(infty)(n dx)/(1+n^(2)x^(2))`, where `a in R,` then L can be

Options

A. `pi`
B. `pi//2`
C. 0
D. 1

Answer

To solve the problem, we need to evaluate the limit:

\[ L = \lim_{n \to \infty} \int_{a}^{\infty} \frac{n \, dx}{1 + n^2 x^2} \]

### Step 1: Rewrite the Integral
We start by rewriting the integral:

\[ I = \int_{a}^{\infty} \frac{n \, dx}{1 + n^2 x^2} \]

### Step 2: Substitution
Next, we can use a substitution to simplify the integral. Let:

\[ u = n x \quad \Rightarrow \quad dx = \frac{du}{n} \]

When \( x = a \), \( u = n a \) and when \( x \to \infty \), \( u \to \infty \). Thus, the limits of integration change accordingly.

Substituting into the integral gives:

\[ I = \int_{n a}^{\infty} \frac{n \cdot \frac{du}{n}}{1 + u^2} = \int_{n a}^{\infty} \frac{du}{1 + u^2} \]

### Step 3: Evaluate the Integral
The integral \( \int \frac{du}{1 + u^2} \) is a standard integral which evaluates to:

\[ \tan^{-1}(u) \]

Thus, we have:

\[ I = \left[ \tan^{-1}(u) \right]_{n a}^{\infty} = \tan^{-1}(\infty) - \tan^{-1}(n a) \]

### Step 4: Calculate the Limits
Since \( \tan^{-1}(\infty) = \frac{\pi}{2} \), we can write:

\[ I = \frac{\pi}{2} - \tan^{-1}(n a) \]

### Step 5: Substitute Back into the Limit
Now we substitute this back into our limit:

\[ L = \lim_{n \to \infty} \left( \frac{\pi}{2} - \tan^{-1}(n a) \right) \]

### Step 6: Evaluate the Limit
As \( n \to \infty \), \( \tan^{-1}(n a) \) approaches \( \frac{\pi}{2} \) (since \( a \) is a constant and \( n \) is going to infinity). Therefore:

\[ L = \frac{\pi}{2} - \frac{\pi}{2} = 0 \]

### Conclusion
Thus, the value of \( L \) is:

\[ L = 0 \]