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

  1. $e^{-1}$
  2. $e^{-1/2}$
  3. $e^{-2}$
  4. $1$
in Calculus by Veteran (431k points)
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1 Answer

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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.
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