Question

The value of $\underset{x\to \infty }{lim}\frac{(\ln x{)}^2}{x^n},n>0$ is equal to

• A. $1$
• B. $0$
• C. $-1$
• D. $\infty$

Answer 1

The correct option is B $0$

Clearly form of the limit is, $\frac{\infty }{\infty }$
Thus applying L-Hospital's rule,

$\underset{x\to \infty }{lim}\frac{(\ln x{)}^2}{x^n}=\underset{x\to \infty }{lim}\frac{2\ln x.\left(\frac{1}{x}\right)}{n{x}^{n-1}}$

$=\frac{2}{n}\underset{x\to \infty }{lim}\frac{\ln x}{x^n},$ again $\frac{\infty }{\infty }$

Thus applying L-Hospital's rule,

$=\frac{2}{n^2}\underset{x\to \infty }{lim}\frac{1}{x{x}^{n-1}}$

$=\frac{2}{n^2}\underset{x\to \infty }{lim}\frac{1}{x^n}=0$