Question

$\underset{x\to \infty }{lim}\frac{lo{g}_{\epsilon }\left[x\right]}{x},\text{where[x]denotes the greatest integer less than or equal to x, is:}$

• A. 1
• B. -1
• C. 0
• D. Does not exist

Answer 1

The correct option is C

0

$\text{Let x=n+k, n}\in /.0\le \text{k}<1;\text{as}x\to \infty ,n\to \infty \text{Then}\underset{x\to \infty }{lim}\frac{lo{g}_e\left[x\right]}{x}=\underset{n\to \infty }{lim}\frac{lo{g}_e[n+k]}{(n+k)}=\underset{n\to \infty }{lim}\frac{lo{g}_{e}n}{(n+k)}\left(\frac{\infty }{\infty }form\right)=\underset{n\to \infty }{lim}\frac{1/n}{1}=0\text{using L-Hospital's rule}$