Question

$\underset{x\to 0}{lim}{\left(\frac{1^{1/x}+2^{1/x}+3^{1/x}+.....n^{1/x}}{n}\right)}^{nx},nϵN$ is equal to

• A. $n!$
• B. $1$
• C. $\frac{1}{n!}$
• D. $0$

Answer 1

The correct option is A $n!$

$\underset{x\to \infty }{lim}{\left(\frac{1^{1/x}+2^{1/x}+3^{1/x}......+n^{1/x}}{n}\right)}^{nx}$

if we put the limits then we have $1^{\infty }$ form. We have for $\underset{x\to \infty }{lim}{\left[f\left(x\right)\right]}^{9\left(x\right)}$ such that limit has the form $1^{\infty }$, then

$\underset{x\to \infty }{lim}{\left[f\left(x\right)\right]}^{9\left(x\right)}=e^{\underset{x\to \infty }{lim}9\left(x\right)[f\left(x\right)-1]}⟶\left(1\right)$

Using this we can evalute the limit asked,

$\underset{x\to \infty }{lim}{\left(\frac{1^{1/x}+2^{1/x}+3^{1/x}+......n^{1/x}}{n}\right)}^{nx}=e^{\underset{x\to \infty }{lim}nx\left(\frac{1^{1/x}+2^{1/x}+3^{1/x}......n^{1/x}-n}{n}\right)}$

$=e^{\underset{x\to \infty }{lim}\left(\frac{1^{1/x}+2^{1/x}+3^{1/x}+......n^{1/x}-n}{\left(1/x\right)}\right)}$

if we put the limits we find that it is in $÷$ form, so we will employ the L'Hospital's rule

$=e^{\underset{x\to \infty }{lim}\left(\frac{1^{1/x}(-\frac{1}{x^2})ln1+2^{1/x}(-\frac{1}{x^2})ln\mathrm{2......}+n^{1/x}(-\frac{1}{x^2})lnn-0}{(-1/x^2)}\right)}$

$=e^{\underset{x\to \infty }{lim}(1^{1/x}ln1+2^{1/x}ln\mathrm{2......}+n^{1/x}lnn)}$

$=e^{ln1+ln2+ln\mathrm{3......}+lnn}$ [$\because$ $\underset{x\to \infty }{lim}{n}^{1/x}=n^0=1$]

$=e^{ln\left(\mathrm{1.2.3.4.5.6......}n\right)}$ [$\because$ $lna+lnb=lnab$]

$=e^{lnn!}$

$=n!$ [$\because$ $e^{lnb}=b$]

Answer : Option A