`lim_(n rarr oo) [(1)/(sqrt""(n^(2)-1)) + (1)/(sqrt""(n^(2) - 2^(2)))+ ….+ (1)/sqrt(([n^(2)- (n-1)^(2)]))]`= ….

Options

A. 0
B. `(pi)/(2)`
C. `pi`
D. none

Answer

To solve the limit

\[ \lim_{n \to \infty} \left[ \frac{1}{\sqrt{n^2 - 1}} + \frac{1}{\sqrt{n^2 - 2^2}} + \ldots + \frac{1}{\sqrt{n^2 - (n-1)^2}} \right], \]

we can break down the solution step by step.

### Step 1: Rewrite the expression
We can express the limit as a summation:

\[ \lim_{n \to \infty} \sum_{k=1}^{n-1} \frac{1}{\sqrt{n^2 - k^2}}. \]

### Step 2: Factor out \(n^2\) from the square root
Notice that we can factor \(n^2\) out of the square root:

\[ \sqrt{n^2 - k^2} = \sqrt{n^2(1 - \frac{k^2}{n^2})} = n\sqrt{1 - \frac{k^2}{n^2}}. \]

Thus, we can rewrite the summation:

\[ \sum_{k=1}^{n-1} \frac{1}{\sqrt{n^2 - k^2}} = \sum_{k=1}^{n-1} \frac{1}{n\sqrt{1 - \frac{k^2}{n^2}}}. \]

### Step 3: Change the summation to a Riemann sum
Now we can factor out \( \frac{1}{n} \):

\[ \frac{1}{n} \sum_{k=1}^{n-1} \frac{1}{\sqrt{1 - \frac{k^2}{n^2}}}. \]

As \(n \to \infty\), the term \(\frac{k}{n}\) approaches a continuous variable \(x\) where \(x = \frac{k}{n}\). The summation can be approximated by an integral:

\[ \frac{1}{n} \sum_{k=1}^{n-1} \frac{1}{\sqrt{1 - \left(\frac{k}{n}\right)^2}} \approx \int_0^1 \frac{1}{\sqrt{1 - x^2}} \, dx. \]

### Step 4: Evaluate the integral
The integral

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

is known to equal \(\frac{\pi}{2}\). This is derived from the fact that the integral of \(\frac{1}{\sqrt{1 - x^2}}\) is \(\sin^{-1}(x)\).

### Step 5: Conclusion
Thus, we find that:

\[ \lim_{n \to \infty} \sum_{k=1}^{n-1} \frac{1}{\sqrt{n^2 - k^2}} = \frac{\pi}{2}. \]

Therefore, the final answer is:

\[ \boxed{\frac{\pi}{2}}. \]