- #1
island-boy
- 99
- 0
hello, I need the proof to show that:
[tex]
e^x = \lim_{n\rightarrow\infty} (1+x/n)^n
[/tex]
Here's what I was able to come up with so far:
since the derivative of [tex]e^x[/tex] is also [tex]e^x[/tex],
then let f(x) = [tex]e^x[/tex]
thus:
D(f(x)) = [tex]\lim_{n\rightarrow\infty} \frac{f(x+h) - f(x)}{h} = \lim_{n\rightarrow\infty} \frac{e^{x+h} - e^x}{h} = e^x\lim_{n\rightarrow\infty} \frac{e^h - 1}{h}[/tex]
so for the derivative of [tex]e^x[/tex] to equal itself,
[tex]\lim_{n\rightarrow\infty} \frac{e^h - 1}{h} = 1[/tex]
so for small values of h, we can write:
[tex]e^h - 1 = h[/tex]
and so
[tex] e = (1+h)^{1/h}[/tex]
Replacing h by 1/n, we get:
[tex] e = (1 + 1/n)^n[/tex]
As n gets larger and approaches infinity, we get:
[tex]e = \lim_{n\rightarrow\infty} (1+1/n)^n[/tex]
so, how do I get:
[tex]
e^x = \lim_{n\rightarrow\infty} (1+x/n)^n
[/tex]
Also, is it true that:
[tex](1+x/n)^n \leq e^x[/tex] and
[tex](1-x/n)^n \leq e^{-x}[/tex] for every natural n and every x element of X?
How can I prove this?
thanks!
[tex]
e^x = \lim_{n\rightarrow\infty} (1+x/n)^n
[/tex]
Here's what I was able to come up with so far:
since the derivative of [tex]e^x[/tex] is also [tex]e^x[/tex],
then let f(x) = [tex]e^x[/tex]
thus:
D(f(x)) = [tex]\lim_{n\rightarrow\infty} \frac{f(x+h) - f(x)}{h} = \lim_{n\rightarrow\infty} \frac{e^{x+h} - e^x}{h} = e^x\lim_{n\rightarrow\infty} \frac{e^h - 1}{h}[/tex]
so for the derivative of [tex]e^x[/tex] to equal itself,
[tex]\lim_{n\rightarrow\infty} \frac{e^h - 1}{h} = 1[/tex]
so for small values of h, we can write:
[tex]e^h - 1 = h[/tex]
and so
[tex] e = (1+h)^{1/h}[/tex]
Replacing h by 1/n, we get:
[tex] e = (1 + 1/n)^n[/tex]
As n gets larger and approaches infinity, we get:
[tex]e = \lim_{n\rightarrow\infty} (1+1/n)^n[/tex]
so, how do I get:
[tex]
e^x = \lim_{n\rightarrow\infty} (1+x/n)^n
[/tex]
Also, is it true that:
[tex](1+x/n)^n \leq e^x[/tex] and
[tex](1-x/n)^n \leq e^{-x}[/tex] for every natural n and every x element of X?
How can I prove this?
thanks!
Last edited: