ISI2016-DCG-56

Question

$\underset{x\rightarrow \infty}{\lim} \left(1+\dfrac{1}{x^{2}}\right)^{x}$ equals

• A. $-1$
• B. $0$
• C. $1$
• D. Does not exist

Answer 1

let us assume,

$y=\lim_{n->\infty}(1+\frac{1}{n^{2}})^{n}$

taking log on both sides,

$\ln y=\lim_{n->\infty}n\ln (1+\frac{1}{n^{2}})$

let take,

$z=\frac{1}{n}$ ,so as,$n-> \infty$ then $z-> 0$

$\ln y=\lim_{z->0}\frac{ln(1+z^{2})}{z}$

$\ln y=\lim_{z->0}z*\frac{ln(1+z^{2})}{z^{2}}$

let assume, $w=z^{2}$ so as z->0 so w->0.

$\ln y=\lim_{z->0} z * \lim_{w->0} \frac{\ln (1+w)}{w}$

$\ln y=0*1=0$

as ,$\lim_{w->0} \frac{\ln (1+w)}{w}=1$

$\lim_{z->0} z=0$

so,$\ln y=0$

y=1.

so correct answer is C.