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Proof of a limit

  1. Oct 29, 2007 #1
    can somebody explain to me how to prove this:

    the limit, as h approaches 0, of ((e^h)-1)/h equals 1.

    sorry about the typing, idk how else to get it on here...
  2. jcsd
  3. Oct 29, 2007 #2
    Do you want a real proof? Or just how you get the answer?

    use L'h rule you get e^h/1 take the limit and you get 1

    but as for a formal proof, i'm to tired to think lol.
  4. Oct 29, 2007 #3
    L'hopitals rule is not a way to do this because to do that one usually has to know the derivative of e^x, but this limit is, itself, needed in order to figure out this derivative. So for a better proof try using the definition of e, this is not particularly hard if you use the definition of e as the limit (1+h)^(1/h) as h approaches zero.
  5. Oct 29, 2007 #4
    can u explain it a lil further? what do you mean by the definition of e?
  6. Oct 29, 2007 #5
    Can you explain it to me? What do you mean when you use e? It is obviously a number, but usually it is defined in some way when one is first introduced to it, in your course or textbook or whatever how is it defined?
  7. Oct 29, 2007 #6
    I think he wants you to look at it this way: [tex]e^h=\sum_{j=0}^\infty\frac{h^j}{j!}=1+h+\frac{h^2}{2!} ++++[/tex] Now plug that into your original equation.
    Last edited: Oct 29, 2007
  8. Oct 29, 2007 #7
    Actually I was aiming for e= the limit as n approaches infinity of [1+(1/n))]^n which is equivalent to the limit as h approaches 0 of (1+h)^(1/h)
  9. Oct 30, 2007 #8
    Well, regardless, if you can make that work out to something like: 1+1+(1-h)/2! +(1-h)(1-2h)/3!+++ all you arrive at as h goes to zero is the value for e, not the value for e^h.

    Now when I took the calculus, I believe e^x was defined as [tex]lim_{n\rightarrow\infty} (1+x/n)^n [/tex]
  10. Oct 30, 2007 #9
    All you need is the value of e though.
  11. Oct 30, 2007 #10
    You mean [tex]\frac{[(1+h)^{1/h}]^h-1}{h}[/tex]?
    Last edited: Oct 30, 2007
  12. Oct 30, 2007 #11
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