Question

Evaluate: $\underset{n\to \infty }{lim}\{{\log }_{n-1}^n.{\log }_n^{n+1}.{\log }_{n+1}^{n+2}\cdots \cdots \cdots \cdots \cdot \cdot {\log }_{n^k-1}^{n^k}\}$

• A. $2n$
• B. $n$
• C. $k$
• D. $2k$

Answer 1

Answer: (C)

$k$

Given: $\underset{n\to \infty }{lim}\{{\log }_{n-1}^n.{\log }_n^{n+1}.{\log }_{n+1}^{n+2}\cdots \cdots \cdots \cdots \cdot \cdot {\log }_{n^k-1}^{n^k}\}$

${\log }_{n-1}^n\cdot {\log }_n^{n+1}=\frac{\log n}{\log (n-1)}\cdot \frac{\log (n+1)}{\log n}={\log }_{n-1}^{n+1}$......simultaneously this form will get reduced to

$=\underset{n\to \infty }{lim}\left(\frac{k\log n}{\log (n-1)}\right)$

Applying L'Hopital's rule

$={lim}_{n\to \infty }\left(\frac{k\frac{1}{n}}{\frac{1}{n-1}}\right)$
$={lim}_{n\to \infty }\left(\frac{k(n-1)}{n}\right)$

$=k$

Hence, option ${}^{\prime }{C}^{\prime }$ is correct.