Question

The value of ${lim}_{x\to 0}{x}^m{\left(\ln x\right)}^n,m,n\in N$ is

• A. $0$
• B. $\frac{m}{n}$
• C. $mn$
• D. None of these

Answer 1

Answer: (A)

$0$

Given $\underset{x\to 0}{lim}{x}^m{\left(\ln x\right)}^n=\underset{x\to 0}{lim}\frac{{\left(\ln x\right)}^n}{x^{-m}}$ $(\frac{\infty }{\infty }$from$)$

$=\underset{x\to 0^{+}}{lim}\frac{n{\left(\ln x\right)}^{n-1}\frac{1}{x}}{{-mx}^{-m-1}}$ (By L.Hospital's Rule)

$=\underset{x\to 0^{+}}{lim}\frac{n{\left(\ln x\right)}^{n-1}}{{-mx}^{-m}}$ $(\frac{\infty }{\infty }$ from$)$

$=\underset{x\to 0^{+}}{lim}\frac{n{(n-1)\left(\ln x\right)}^{n-2}\frac{1}{x}}{{(-m)}^2{x}^{-m-1}}$ (By L-Hospital's Rule)

$=\underset{x\to 0^{+}}{lim}\frac{n(n-1){\left(\log x\right)}^{n-2}}{m^2{x}^{-m}}$ (Diff. numerator and denominator $n$ times)

$=\underset{x\to 0^{+}}{lim}\frac{n!}{{(-m)}^n{x}^{-m}}=0$