ANOTHER really basic question this time regarding e

[Mathguy15]

ANOTHER really basic question... this time regarding e :D

Hello,

As some of you well know, I'm reading through some lecture notes on single-variable calculus. The teacher who wrote these notes gives a proof of exp'(x)=exp(x). He wrote this inequality:

[((1+1/n)^n)-1]n ≤ [(e^h)-1]/h≤ [((1+1/n)^n+1)-1]n

Where h=1/n.

Now, I get why this inequality is true, but the teacher says that the right and left sides go to 1 as n--->∞. He said to use the binomial expansions to see this. I'm not getting it. Could someone help?

Thanks(again)
mathguy

 

[Mandlebra]

Use the binomial formula on (1+ 1/n)^ n

 

[Mathguy15]

Ugh, I tried.. I still don't get the intuition.

In fact, I have this intuition:

as n--->∞, (1+1/n)^n --->e, which is 2.718blah blah blah. This minus 1 is 1.718blahblahblah. This multiplied by a large number would be very far from 1. Is there something I'm missing?

 

[Millennial]

I would go a bit different about the limit, it is easy to put the derivative after the limit, differentiate, and see that you get the same expression. But, as far as I saw, your inequality can be written more clearly as follows:
$$\left(\left(1+\frac{1}{n}\right)^n-1\right)n\leq n(e^{1/n}-1)\leq \left(\left(1+\frac{1}{n}\right)^{n+1}-1\right)n$$
If you want to use this, you need to expand out the binomial and remember that one over n goes to zero as n goes to infinity.

 

[dextercioby]

I don't see how the binomial thing would help. The proof for derivative is actually for

$$\lim_{h\rightarrow 0} \frac{e^h - 1}{h} = 1$$