gate 2010

Question

What is the value of Lim_n->∞(1-1/n)²ⁿ ?
(A) 0
(B) e^-2
(C) e^-1/2
(D) 1

Answer 1

$f\left ( x \right ) = (1 - \frac{1}{n})^{2n}$

f(x) is of the form $1^{\infty }$ 

Taking y = f(x),

$\ln y = 2n*\lim_{n \to \infty }\ln (1-\frac{1}{n})$

$\Rightarrow $$\ln y = 2* \lim_{n \to \infty }\frac{\ln ({1 - \frac{1}{n})}}{1/n}$    ($\frac{0}{0}$ form)

Applying L'Hospitals rule,

$\ln y = 2* \lim_{n \to \infty }(\frac{\frac{1}{1 - \frac{1}{n}}*(-\frac{1}{n^{2}})}{-1/n^{2}})$

$\Rightarrow \ln y = -2*\lim_{n \to \infty }(\frac{1}{1-\frac{1}{n}})$

$\Rightarrow \ln y = -2$

 $\Rightarrow y = e^{-2}$

Hence option B.