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The limit $\lim \limits_{n \to \infty} \bigg(1-\dfrac{1}{n^2}\bigg)^{n}$ equals?

$(A) e^{-1}$

$(B) e^{-\frac{1}{2}}$

$(C) e^{-2}$

$(D) 1$

Using the expansion $(1+x)^{n}=1+nx+ \dfrac{n(n-1)x^{2}}{2!} ...$
I am having large degree of $n$ in denominator therefor i am left with $1$ after applying the limit. Did i do it right ? If i did it right is there any alternative method to solve this problem like a quick version?

Martin Sleziak
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Daman
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4 Answers4

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

  • It looks so easy now. How did miss that anyways thanks:) – Daman Feb 05 '18 at 11:12
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    @Damn1o1 I don't know if you were supposed to solve the problem that way, but at least it serves as a good way to check your result that you may have obtained differently. –  Feb 05 '18 at 11:16
  • No it was best I am building up speed for MCQ exam i shouldn't miss these types of catch. I guess they come with experience. – Daman Feb 05 '18 at 11:18
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$1-\frac{1}{n^2} \to 1$ and $n \to \infty$, so: $$\lim_{n \to \infty}\left(1-\frac{1}{n^2}\right)^n=\exp\left(\lim_{n \to \infty}\left(1-\frac{1}{n^2}-1\right)n\right)=\exp\left(\lim_{n \to \infty}\left(-\frac{1}{n^2}\right)n\right)=\exp\left(\lim_{n \to \infty}-\frac{1}{n}\right)=\exp(-0)=\exp(0)=1$$

Botond
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Hint:

$$\lim_{n\to\infty}\left(1-\dfrac1{n^2}\right)^n=\left(\lim_{n\to\infty}\left(1+\dfrac1{-n^2}\right)^{-n^2}\right)^{\lim_{n\to\infty}-1/n}$$

Set $x=1$ in About $\lim \left(1+\frac {x}{n}\right)^n$

lab bhattacharjee
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By Bernoulli: $ \left(1-\dfrac1{n^2}\right)^n \ge 1-\frac{1}{n}$, hence

$$1-\frac{1}{n} \le \left(1-\dfrac1{n^2}\right)^n \le 1.$$

Conclusion ?

Fred
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