If `|cos^-1(1)/(n)|lt (pi)/(2)`, then
`lim_(n to oo) {(n+1)(2)/(pi)cos^-1.(1)/(n)-n}`

Answer

To solve the limit problem given by the expression:

\[ \lim_{n \to \infty} \left( (n + 1) \frac{2}{\pi} \cos^{-1} \left( \frac{1}{n} \right) - n \right) \]

we will follow these steps:

### Step 1: Rewrite the limit expression
We can factor out \(\frac{2}{\pi}\) from the limit:

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

### Step 2: Simplify the expression inside the limit
Now, we can expand the expression:

\[ (n + 1) \cos^{-1} \left( \frac{1}{n} \right) = n \cos^{-1} \left( \frac{1}{n} \right) + \cos^{-1} \left( \frac{1}{n} \right) \]

Thus, the limit becomes:

\[ \lim_{n \to \infty} \frac{2}{\pi} \left( n \cos^{-1} \left( \frac{1}{n} \right) + \cos^{-1} \left( \frac{1}{n} \right) - \frac{\pi n}{2} \right) \]

### Step 3: Use the identity for \(\cos^{-1}\)
We know that:

\[ \cos^{-1} \left( \frac{1}{n} \right) = \frac{\pi}{2} - \sin^{-1} \left( \frac{1}{n} \right) \]

Substituting this into our limit gives:

\[ \lim_{n \to \infty} \frac{2}{\pi} \left( n \left( \frac{\pi}{2} - \sin^{-1} \left( \frac{1}{n} \right) \right) + \left( \frac{\pi}{2} - \sin^{-1} \left( \frac{1}{n} \right) \right) - \frac{\pi n}{2} \right) \]

### Step 4: Simplify further
This simplifies to:

\[ \lim_{n \to \infty} \frac{2}{\pi} \left( n \frac{\pi}{2} - n \sin^{-1} \left( \frac{1}{n} \right) + \frac{\pi}{2} - \sin^{-1} \left( \frac{1}{n} \right) - \frac{\pi n}{2} \right) \]

The \(n \frac{\pi}{2}\) and \(-\frac{\pi n}{2}\) cancel out, leading to:

\[ \lim_{n \to \infty} \frac{2}{\pi} \left( -n \sin^{-1} \left( \frac{1}{n} \right) + \frac{\pi}{2} - \sin^{-1} \left( \frac{1}{n} \right) \right) \]

### Step 5: Evaluate the limit
As \(n \to \infty\), \(\sin^{-1} \left( \frac{1}{n} \right) \approx \frac{1}{n}\). Thus,

\[ -n \sin^{-1} \left( \frac{1}{n} \right) \approx -n \cdot \frac{1}{n} = -1 \]

So we have:

\[ \lim_{n \to \infty} \frac{2}{\pi} \left( -1 + \frac{\pi}{2} - 0 \right) = \frac{2}{\pi} \left( \frac{\pi}{2} - 1 \right) = 1 - \frac{2}{\pi} \]

### Final Answer
Thus, the limit evaluates to:

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