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# KCC's quizzes AQQ255 about a limit at infinite

And here is a new quiz coming!

1. Info: the frequent winners should have received some award and certificate. Please warn me if you think you have been forgotten!

2. Quote of the week: "The elevator to success is out of order. You'll have to use the stairs... one step at a time." - Joe Girard.

3. New quiz of the week: AQQ255 about a limit to infinite

Good luck!

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[edited by: emassa at 3:52 PM (GMT -4) on 21 Mar 2024]
• Yes! You applied the right approach to prove it in a very formal way! Congratulation  ! The Hospital method saying Lim[(Fx)/G(x)] = Lim[(F’x)/G’x)] gives quickly the solution.

• Ok, game over! Here is the official answer:

Solution:

1. The limit is the neperian number e. Thus close to 2.71One can guess that number by using a simple calculator having the exponent function. With x=1000, you obtain already 2.7169And by increasing x to 10000, then 10000 you see the value converging to 2.7182804 which is very close to e (=2.7182818...). Of course one needs to remember e value is like that...
2. To prove it, let’s call k = Lim(1+1/x))x

Ln(k) = Lim[x.Ln(1+1/x)] for x going to infinite

Let’s now call 1/x = y and thus x=1/y

The limit can be rewritten as:

Ln(k)        = Lim((1/y). Ln(1+y) for y going to 0

= Lim(Ln(1+y)/y) for y going to 0

By using the Hospital rule saying Lim[(Fx)/G(x)] = Lim[(F’x)/G’x)]

We can write Ln(k) = 1/(1+y) for y going to 0

Thus Ln(k) = 1 and thus k = e

Funny isnt' it?

So, congratulations to the ones having found the solution and the way to proove it (without Excel)!

Big applause to our 4 winners:

1. Barry Kulp, Test Engineer Retired from Maxim, Beaverton, USA

2. Tim Watkins, Application Engineer ADI, USA