ISI2015-MMA-20

Question

The limit $\underset{n \to \infty}{\lim} \left( 1- \frac{1}{n^2} \right) ^n$ equals

• A. $e^{-1}$
• B. $e^{-1/2}$
• C. $e^{-2}$
• D. $1$

Answer 1

Answer $D$

Given:$$\underset{n\to\infty}{\lim}{\Big(1-\frac{1}{n^2}\Big)}^n\qquad \to (1)$$

The above can be written as:

$$\underset{n\to\infty}{\lim}{\Big(1-\frac{1}{n}\Big)}^n\underset{n\to\infty}{\lim}{\Big(1+\frac{1}{n}\Big)}^n\qquad \to (2)$$, using the identity$a^2-b^2 = (a+b)(a-b)$

Now, we know that:

$$\underset{n\to\infty}{\lim}{\Big (1-\frac{1}{n}\Big)}^n = \frac{1}{e} \qquad \to (3),\;and\; \underset{n\to\infty}{\lim}{\Big (1+\frac{1}{n}\Big) }^n= e\qquad \to (4)$$

From equations $(1), \;(2) \;and (3)$, we get:

$$\underset{n\to\infty}{\lim}{\Big(1-\frac{1}{n^2}\Big)}^n = \frac{1}{e}\times e = 1$$

$\therefore D$ is the correct option.

Answer 2

Answer: D

--------------------------------------------------------------------

An easy and standard 12^th class method:

If $$\lim_{x\to a} f(x)=1$$ and  $$\lim_{x\to a} g(x)=+\infty$$ then $$\lim_{x\to a}f^g=e^{\lim_{x\to a}(f-1)g}$$

 

Here $f(n)=1-\frac{1}{n^2}$ and $g(n)=n$. Therefore,

$$\lim_{n\to\infty}(1\frac{-1}{n^2}-1)n=0$$

Thus, $$\lim_{n\to\infty}\left(1-\frac{1}{n^2}\right)^{n}=e^0=1$$

Answer 3

let assume ,

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

taking log both side,

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

let us assume z=1/n , so as $n->\infty$ ,z->0.

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

it is now is in $\frac{0}{0}$ form , so we can use L’Hopital rule,

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

$\ln y=0$

$y=1$

so correct is (D).