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!
I would've taken the expression for e and raised both sides to the x power. You really ought to do this more rigorously... P.S. many of the limits in your limits are wrong.
which one of the limits are wrong? I was able to prove that: [tex] (1 + x/n)^n \leq e^x [/tex] and [tex] (1 - x/n)^n \leq e^{-x} [/tex] Using the Taylor expansion for [tex]e^x[/tex], I got [tex]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...[/tex] which means [tex]e^x \geq 1 + x[/tex] subtituting x by x/n, I get [tex]e^{x/n} \geq 1 + x/n[/tex] thus [tex]e^{(x/n)n} \geq (1 + x/n)^n[/tex] simplifying, I get: [tex]e^x \geq (1 + x/n)^n [/tex] Similarly I used the Taylor Expnansion to prove that [tex] e^x = \lim_{n\rightarrow\infty} (1+x/n)^n [/tex] since [tex]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +...[/tex] and using binomial theorem for [tex]\lim_{n\rightarrow\infty} (1+x/n)^n[/tex] I got: [tex]\lim_{n\rightarrow\infty}1^n + \lim_{n\rightarrow\infty} x + \lim_{n\rightarrow\infty} \frac{n(n-1)(x^2)}{(n^2)(2!)} + \lim_{n\rightarrow\infty} \frac{n(n-1)(n-2)(x^3)}{(n^3)(3!)} + \lim_{n\rightarrow\infty} \frac{n(n-1)(n-2)(n-3)(x^4)}{(n^4)(4!)} + ... = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +... =e^x [/tex] these correct?
See this thread posts # 3 and 12. I take it that your definition for [tex]e^x[/tex] is the Taylor series about x=0, viz. [tex]e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}[/tex]