`lim_(n to infty) log_(n-1)(n) log_(n)(n+1)….log_(n^(k)-1) (n^(k))` is equal to:

Options

A. k
B. 2k
C. 3k
D. 4k

Answer

To solve the limit

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

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 can express each logarithm in the product as follows:

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

### Step 2: Write the product of logarithms

The limit can now be rewritten as:

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

### Step 3: Simplify the product

Notice that in this product, most terms will cancel out. Specifically, we can see that:

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

### Step 4: Analyze the limit as \( n \to \infty \)

As \( n \to \infty \):

- \(\log(n) \to \infty\)
- \(\log(n-1) \to \infty\)
- \(\log(n+1) \to \infty\)
- \(\log(n^{k}) = k \log(n) \to \infty\)
- \(\log(n^{k}-1) \to \infty\)

Thus, we are dealing with an indeterminate form of type \(\frac{\infty}{\infty}\).

### Step 5: Apply L'Hôpital's Rule

To resolve the indeterminate form, we apply L'Hôpital's Rule. We differentiate the numerator and denominator:

1. Differentiate the numerator: \(\frac{d}{dn} \log(n^{k}) = \frac{k}{n}\).
2. Differentiate the denominator: \(\frac{d}{dn} \log(n-1) = \frac{1}{n-1}\).

Now we can write:

\[ \lim_{n \to \infty} \frac{k/n}{1/(n-1)} = \lim_{n \to \infty} \frac{k(n-1)}{n} = k \cdot \lim_{n \to \infty} \left(1 - \frac{1}{n}\right) = k. \]

### Final Result

Thus, we conclude that:

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