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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]
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• 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

4. Sparky, engineer retired from Lockheed Martin, USA

Now, be prepared for our next coming challenge!

• 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