`lim_(n rarr infty) sum_(k=1)^(n) (n)/(n^(2)+k^(2)x^(2)),x gt 0 ` is equal to

Options

A. `x tan ^(-1)(x)`
B. `tan^(-1)(x)`
C. `(tan^(-1)(x))/(x)`
D. `tan^(-1)(x)/(x)`

Answer

To solve the limit problem, we start with the expression:

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

### Step 1: Rewrite the Sum
We can rewrite the sum by factoring out \( n \) from the denominator:

\[ \sum_{k=1}^{n} \frac{n}{n^2 + k^2 x^2} = \sum_{k=1}^{n} \frac{n}{n^2(1 + \frac{k^2 x^2}{n^2})} = \sum_{k=1}^{n} \frac{1}{n(1 + \frac{k^2 x^2}{n^2})} \]

### Step 2: Change of Variable
Now, we can express this sum in terms of \( t = \frac{k}{n} \). Thus, \( k = nt \) and \( \Delta t = \frac{1}{n} \):

\[ \sum_{k=1}^{n} \frac{1}{n(1 + \frac{k^2 x^2}{n^2})} = \sum_{k=1}^{n} \frac{1}{n(1 + t^2 x^2)} \cdot \Delta t \]

### Step 3: Convert to Integral
As \( n \to \infty \), the sum approaches the integral:

\[ \int_{0}^{1} \frac{1}{1 + t^2 x^2} \, dt \]

### Step 4: Evaluate the Integral
The integral can be evaluated using the formula for the integral of \( \frac{1}{1 + a^2 t^2} \):

\[ \int \frac{1}{1 + a^2 t^2} \, dt = \frac{1}{a} \tan^{-1}(at) + C \]

In our case, \( a = x \):

\[ \int_{0}^{1} \frac{1}{1 + t^2 x^2} \, dt = \frac{1}{x} \left[ \tan^{-1}(xt) \right]_{0}^{1} = \frac{1}{x} \left( \tan^{-1}(x) - \tan^{-1}(0) \right) = \frac{1}{x} \tan^{-1}(x) \]

### Step 5: Final Result
Thus, we have:

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{n}{n^2 + k^2 x^2} = \frac{1}{x} \tan^{-1}(x) \]

### Conclusion
The final answer is:

\[ \frac{1}{x} \tan^{-1}(x) \]

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