Barion Pixel Sequences tending to e | mathXplain
 

Contents of this Calculus 2 episode:

Find the limits, Sequences tending to e, Limits of exponential sequences.

Text of slideshow

Here come some very funny sequences.

Here is the first one.

Well, this is not the most difficult limit of our lives.

Here we have another one.

And a third one.

Well, there is nothing funny so far.

The real thrill comes now.

Here is this limit.

Based on the previous examples, this is supposedly 1, again.

The problem is: it is not.

It is because the exponent is not a specific number, but the index of the sequence, so it changes all the time.

If the exponent is a specific number, then all we did so far is OK.

But if the exponent contains n, then we are dealing with an entirely different animal.

In this case the limit is not 1, but a rather unpleasant number.

To avoid looking at this unpleasant number so often, they called it "e".

If it is not 1 here, but some other number,

then the limit will change a little bit.

And there is one more thing.

At least when

Well, let's see a few examples for this.

Here is this limit, for instance:

which is based on the formula

But if this part here changes,

and so does the exponent,

then, we get the same thing again.

Here is another one:

So, limits where these are the same can be computed very easily.

The question is what happens if they are not the same.

Then, we have to do something to make them the same. We either turn this into 2n,

or this into n.

Let's turn this into n.

Let's simplify the fraction by 2. And now it is good.

Here we will transform the exponent.

There is such a thing as:

There are some cases that are a bit more exciting:

But these are still not as exciting as those that come next.

Limits where appears here

and in the exponent, too,

can always be computed using these formulas.

The trick is to twist and turn things until they resemble these formulas.

Well, to achieve our goal, let's divide the numerator and the denominator by .

If, by any chance, there is in both the numerator and the denominator, then we divide by :

And we can fix uglier cases, too

If the exponent is a specific number, then:

But if, unfortunately, here we have

and there

then we only divide by , and then factor out

Sometimes there is n2 there.

Furthermore, it could be n3.

There is 2n3 at the top and also at the bottom, so let's divide by 2n3.

 

Sequences tending to e

03
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