STD 11 - 12. limits — JEE STD 11 Science — Question

 $= n \pi $ $\left( {1\,\, + \,\,\frac{1}{{2\,n}}\,\, + \,\,\frac{1}{2}\,\left( {\frac{1}{2}\,\, - \,\,1} \right)\,\,\frac{1}{{2\,\,!}}\,\,\frac{1}{{{n^2}}}\,\, + \,\,......} \right)$
$= \pi \left[ {n\,\, + \,\,\frac{1}{2}\,\, + \,\,\frac{1}{2}\,\left( {\frac{1}{2}\,\, - \,\,1} \right)\,\,\frac{1}{{2\,\,!}}\,\,\frac{1}{n}\,\, + \,\,......} \right)$
as $n \Rightarrow \infty$ ;

$\frac{\pi }{2}$. $\left( {2\,n\,\, + \,\,1\,\, + \,\,\left( {\frac{1}{2}\,\, - \,\,1} \right)\,\,\frac{1}{{2\,\,!}}\,\,\frac{1}{n}\,\, + \,\,.......} \right)$ 

$= (2n + 1)$ $\frac{\pi }{2}$
Alternatively  $(1)$ :

Take $A = (2n +1)$ $\frac{\pi }{2}\,$ and $B = $ $\frac{\pi }{2}\,$$\left[ {\left( {\frac{1}{2}\, - \,1} \right)\,\frac{1}{{2!}}\,\,\frac{1}{n}\,\, + \,\,....} \right]$
now $cos (A + B) = cosA cosB - sinA sinB $
$= 0 - (1)$$\mathop {Limit}\limits_{n\,\, \to \,\,\infty } $ $sin$ $\left( {\frac{\pi }{2}\,\left( {\frac{1}{2}\, - \,1} \right)\frac{1}{{2!}}\,\frac{1}{n}\, + \,...} \right)$
$= 0 $
Alternatively $(2)$

Best $l =$ $\mathop {Limit}\limits_{n \to \infty } \,\, \pm \,\,\cos \left( {n\pi \, - \,\pi \sqrt {{n^2} + n} \,} \right)$

$\Rightarrow$ $=\mathop {Limit}\limits_{n \to \infty } \,\, \pm \,\,\cos \left( {\,\pi \,\,\left( {n\, - \,\sqrt {{n^2} + n} } \right)\,} \right)$
$=\mathop {Limit}\limits_{n \to \infty } \,\,\,\cos \,\left( {\frac{{\left( {\pi \,( + n)} \right)}}{{n + \sqrt {{n^2} + n} }}} \right)\,\,$

= $\mathop {Limit}\limits_{n \to \infty } \,\,\,\cos \,\left( {\frac{{n\,\pi }}{{n + n\,\sqrt {1 + \frac{1}{n}} }}} \right)\,\,$

= $\mathop {Limit}\limits_{n \to \infty } \,\,\,\cos \,\left( {\frac{{\,\pi }}{{1 + \,\sqrt {1 + \frac{1}{n}} }}} \right)\,\,$

$= cos$ $\frac{\pi }{2}\,$ $\to 0 $