concept on e and e^x

simplest way to express e^x is
lim(n->oo) (1+x/n)^n
The prove is just a change of variable: let m = n/x.
Then n->00 iff m->oo, since x is assumed here a constant. (We fixed x)
So the limit becomes:
lim(n->oo) (1+x/n)^n
= lim(m->oo) (1+1/m)^(mx)
= e^x

樓上的Taylor's expansion 答得非常精彩
小弟亦有另外一個見解 , 呢個答法唔雖要咁麻煩
其實...
e^x = (1+x/n)^n for n is a integer tends to infinity

We can use the Taylor's series of expansion of function:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Dec07/Crazydiff12.jpg


where c and x are some values.

Now, take c = 0, we have:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Dec07/Crazydiff13.jpg


which is known as the Maclaurin's series (Taylor's series expansion at c = 0)

So for exponential function, we have:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Dec07/Crazydiff14.jpg


And then the different orders of derivatives are

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Dec07/Crazydiff15.jpg


So eventually we have:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Dec07/Crazydiff16.jpg


An example of application is the Euler's form of complex number which is:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Dec07/Crazydiff17.jpg


After comparing the real and imaginary parts, we have:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Dec07/Crazydiff18.jpg

which is consistent with the typical derivation of sine and cosine functions using Taylor's series method.

(Note: θ is measured in radian.)