Question

if $\ell =\underset{x\to \infty }{lim}{\left|\begin{array}{c}\frac{x+1}{x-1}\end{array}\right|}^x$ then,$\left\{\ell \right\}and\left[\ell \right]\left(where\right\{.\}\&[.]$ denotes the fractional part function and greatest integer function respectively), is/are -

• A. $7$
• B. $7-e^2$
• C. $-7$
• D. $e^2-7$

Answer 1

Answer: (A), (D)

The correct options are
A $7$
D $e^2-7$

$l=\underset{x\to \infty }{lim}{\left(\frac{x+1}{x-1}\right)}^x$

$1^{\infty }$ form

use a special function

$l=e^{{lim}_{x\to \infty }\left\{f\right(x)-1\}g\left(x\right)}$

$f\left(x\right)=\frac{x+1}{x-1}$ and $g\left(x\right)=x$

$l=e^{\underset{x\to \infty }{lim}(\frac{x+1}{x-1}-1)\times x}$

$l=e^{\underset{x\to \infty }{lim}(\frac{x+1-x+1}{n-1}-1)\times x}$

$l=e^{2\underset{x\to \infty }{lim}\frac{x}{x-1}}=e^{2\underset{x\to \infty }{lim}\frac{x}{x(1-\frac{1}{x})}}$

$l=e^{2\left(\frac{1}{1-\frac{1}{\infty }}\right)}=e^{2\left(\frac{1}{1-0}\right)}=e^2=7.39$

$\left[l\right]=\left[7.39\right]=7=e^2$

$\left\{l\right\}=l-\left[x\right]=7.39-7=0.39$ OR $e^2-7$

$\frac{\left[l\right]=7}{\left\{l\right\}=0.39}$ OR $\frac{\left[l\right]=7}{\left\{l\right\}=e^2-7}$