Elementary Limit

Calculus Level 3

$\large \lim_{n\to \infty} \left (1-\frac{1}{n^2}\right)^{n} = \, ?$

e^{-1}

e^{-2}

0

1

e^{-1/2}

2 solutions

Guillermo Templado
Aug 6, 2016

$\displaystyle \lim_{n\to\infty} \left(1 - \frac{1}{n^2}\right)^n = \lim_{n\to\infty} \left(1 - \frac{1}{n^2}\right)^{\frac{-n^2}{-n}} =$ $= \lim_{n\to\infty} \left(\left(1 - \frac{1}{n^2}\right)^{-n^2}\right)^{\frac{1}{-n}} = "\lim_{n\to\infty} e^{\frac{1}{-n}}" = 1$

but if we think logically and not just mathematically , 1 - something would be less than 1 and , something less than 1 when multiplied by itself an infinite amount would become zero , plz tell if i am wrong.

A Former Brilliant Member - 4 years, 9 months ago

Yes, I think you are wrong,I think your problem is that these terms approach to $1$ . I can prove that $\displaystyle\lim_{x \to a} (1 + f(x))^{\frac{1}{f(x)}} = e$ , if $\displaystyle \lim_{x \to a} f(x) = 0$ . Do you want me to prove it? It's similar to $\displaystyle \lim_{n \to \infty} (1 + \frac{1}{n})^{n} = e = \lim_{n \to \infty} (1 - \frac{1}{n})^{-n}$ " $= 1^{\infty}$ "

Guillermo Templado - 4 years, 9 months ago

no, i already know that identity but i want to say that 1/n^2 is infintely small but still has some value and when we subtract it from 1 , the number we get will approach 1 from left side on number line . and something even a little less than 1 when raised to power infinity will tend to zero. i also could have reached the answer mathematically but i want to solve it using pure arguments . plz help

A Former Brilliant Member - 4 years, 9 months ago

@A Former Brilliant Member – Sometimes, we are wrong about some things that seem logical and obviously true for us, for instance, $(1 - \frac{1}{n})^{n}$ seems approaching to $0$ , because $(1 - \frac{1}{n}) < 1$ , but actually $\displaystyle\lim_{n \to \infty} (1 - \frac{1}{n})^{n} = \frac{1}{e}$ . Nevertheless, there is always a true and correct justification (logic) demonstrating that we are wrong. If you want I can detail you one concrete question? Do you have one concrete question for me? do you want me to prove(wit logic-mathemathics arguments) that $\displaystyle\lim_{n \to \infty} (1 + \frac{1}{n})^{n} = e \Rightarrow \displaystyle\lim_{n \to \infty} (1 - \frac{1}{n})^{n} = \frac{1}{e}$

Guillermo Templado - 4 years, 9 months ago

@Guillermo Templado – yes i would be glad if you prove it logically only with giving arguments and not some textbook proof, thanks for seeing to my problem

A Former Brilliant Member - 4 years, 9 months ago

@A Former Brilliant Member – The sequence $(1 + 1/n)^{n}$ is a strictly increasing sequence and it's bounded by 3. This implies that this sequence has a limit. If you take a calculator, you'll see that the limit is approaching to $e \approx 2.718...$ . Make the proof. Take again other calculator and make the same with $(1 - 1/n)^{n}$ , it will convince you, little by little to the right result. If you want me to make a logical proof, I'll be using a text book... My advice: You have to convince yourself... Start with a calculator....

Guillermo Templado - 4 years, 9 months ago

Chew-Seong Cheong
Oct 23, 2016

$\begin{aligned} L & = \lim_{n \to \infty} \left(1-\frac 1{n^2} \right)^n \\ & = \lim_{n \to \infty} \exp \left(n \ln \left(1-\frac 1{n^2} \right)\right) & \small {\color{#3D99F6}\exp (x) = e^x} \\ & = \lim_{n \to \infty} \exp \left(\frac {\ln \left(1-\frac 1{n^2} \right)}{\frac 1n} \right) & \small {\color{#3D99F6} \text{0/0 case, L'Hôpital's rule applies}} \\ & = \lim_{n \to \infty} \exp \left(\frac {\frac {2n}{n^2-1}-\frac {2n}{n^2}}{-\frac 1{n^2}} \right) & \small {\color{#3D99F6} \text{Differentiate up and down}} \\ & = \lim_{n \to \infty} \exp \left( - \frac {2n^2}{n(n^2-1)} \right) & \small {\color{#3D99F6} \text{Divide up and down by }n^3} \\ & = \lim_{n \to \infty} \exp \left( - \frac {\frac 2n}{1-\frac 1{n^2}} \right) \\ & = e^0 = \boxed{1} \end{aligned}$

Elementary Limit

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