Question

$\underset{x\to \infty }{lim}\frac{{\log }_e\left[x\right]}{x}$ (where $[.]$ is $G.I.F-$)$ is

• A. $0$
• B. $1$
• C. $8$
• D. $4$

Answer 1

Answer: (A)

$0$

$\underset{x\to \infty }{lim}\frac{lnx}{x}=0$

${\log }_{e}x=lnx$

$\to$ lnx goes to $\infty$ as x goes to $\infty$

$ln\left(\infty \right)=\infty$

$\therefore$ as $x\to \infty \frac{lnx}{x}\to \frac{\infty }{\infty }$, which is undefined

Thus we apply L' Hospital's Rule

$\underset{x\to \infty }{lim}\frac{lnx}{x}=\underset{x\to \infty }{lim}\frac{\left(\frac{d}{dx}\right)lnx}{\left(\frac{d}{dx}\right)\left(x\right)}=\underset{x\to \infty }{lim}\frac{1}{x}=0$

The limit approches 'O' because 1 divided over something approching $\infty$ becomes closer and closer to zero.

Eg: $\frac{1}{10}=0.1,\frac{1}{100}=0.01,\frac{1}{10000}=0.0001$

$\therefore \underset{x\to \infty }{lim}\frac{{\log }_{e}x}{x}=0$