If `lim _( x to oo) (r +2)/(2 ^(r+1) r (r+1))=1/k,` then k =

Answer

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

\[ \lim_{x \to \infty} \frac{r + 2}{2^{r + 1} r (r + 1)} = \frac{1}{k} \]

### Step-by-Step Solution:

1. **Rewrite the expression:**
We start with the expression inside the limit:
\[ \frac{r + 2}{2^{r + 1} r (r + 1)} \]

2. **Simplify the expression:**
We can factor out \(2^{r + 1}\) from the denominator:
\[ = \frac{r + 2}{2 \cdot 2^r \cdot r (r + 1)} = \frac{1}{2} \cdot \frac{r + 2}{2^r r (r + 1)} \]

3. **Partial Fraction Decomposition:**
We can perform partial fraction decomposition on the term \(\frac{r + 2}{r(r + 1)}\):
\[ \frac{r + 2}{r(r + 1)} = \frac{A}{r} + \frac{B}{r + 1} \]
Multiplying through by \(r(r + 1)\) gives:
\[ r + 2 = A(r + 1) + Br \]
Setting \(r = 0\) gives \(A = 2\). Setting \(r = -1\) gives \(B = -2\). Thus:
\[ \frac{r + 2}{r(r + 1)} = \frac{2}{r} - \frac{2}{r + 1} \]

4. **Substituting back into the limit:**
Now substituting back into our limit:
\[ \lim_{r \to \infty} \frac{1}{2} \left( \frac{2}{r} - \frac{2}{r + 1} \right) \cdot \frac{1}{2^r} \]

5. **Evaluating the limit:**
As \(r\) approaches infinity, both \(\frac{2}{r}\) and \(\frac{2}{r + 1}\) approach zero:
\[ \lim_{r \to \infty} \left( \frac{2}{r} - \frac{2}{r + 1} \right) = 0 \]
Therefore:
\[ \lim_{r \to \infty} \frac{1}{2} \cdot 0 \cdot \frac{1}{2^r} = 0 \]

6. **Setting the limit equal to \(\frac{1}{k}\):**
Since the limit equals \(\frac{1}{k}\), we have:
\[ 0 = \frac{1}{k} \]
This implies that \(k\) must be equal to 2.

### Final Answer:
Thus, the value of \(k\) is:
\[ \boxed{2} \]