`lim_(n->oo)[log_(n-1)(n)log_n(n+1)*log_(n+1)(n+2).....log_(n^k-1) (n^k)]` is equal to :

Options

A. `oo`
B. n
C. k
D. none of these

Answer

To solve the limit

\[ \lim_{n \to \infty} \left[ \log_{n-1}(n) \log_n(n+1) \log_{n+1}(n+2) \cdots \log_{n^k-1}(n^k) \right], \]

we will break it down step by step.

### Step 1: Rewrite the logarithms in terms of natural logarithms

Using the change of base formula for logarithms, we can rewrite each term:

\[ \log_{a}(b) = \frac{\log(b)}{\log(a)}. \]

Thus, we have:

\[ \log_{n-1}(n) = \frac{\log(n)}{\log(n-1)}, \quad \log_n(n+1) = \frac{\log(n+1)}{\log(n)}, \quad \log_{n+1}(n+2) = \frac{\log(n+2)}{\log(n+1)}, \ldots, \quad \log_{n^k-1}(n^k) = \frac{\log(n^k)}{\log(n^k-1)}. \]

### Step 2: Write the product

The entire expression can be rewritten as:

\[ \frac{\log(n)}{\log(n-1)} \cdot \frac{\log(n+1)}{\log(n)} \cdot \frac{\log(n+2)}{\log(n+1)} \cdots \frac{\log(n^k)}{\log(n^k-1)}. \]

Notice that in this product, many terms will cancel out:

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

### Step 3: Simplify the logarithm

We know that:

\[ \log(n^k) = k \log(n). \]

Thus, we can rewrite the limit as:

\[ \lim_{n \to \infty} \frac{k \log(n)}{\log(n-1)}. \]

### Step 4: Analyze the limit

As \( n \to \infty \), we can approximate \(\log(n-1)\):

\[ \log(n-1) \sim \log(n) \quad \text{(since } n-1 \text{ approaches } n \text{ as } n \to \infty\text{)}. \]

So we have:

\[ \lim_{n \to \infty} \frac{k \log(n)}{\log(n-1)} = \lim_{n \to \infty} \frac{k \log(n)}{\log(n)} = k. \]

### Final Result

Thus, the limit evaluates to:

\[ \lim_{n \to \infty} \left[ \log_{n-1}(n) \log_n(n+1) \log_{n+1}(n+2) \cdots \log_{n^k-1}(n^k) \right] = k. \]