Taylor series of $e^x$

Question

When I have something like $$\lim _{x\to \:0}\left(\frac{1}{e^x}\right)$$ and I want to write its Taylor polynomial, I just use $$\lim _{x\to \:0}\left(\frac{1}{e^x}\right) = \lim _{x\to \:0}\left(e^{-x}\right) = 1\:-\:x\:+\:x^2/2\:-\:x^3/6\:+\:x^4/24\:-\:x^5/120\:+ ....$$ But how can that be to just expanding Taylor at denominator, like $$\lim _{x\to \:0}\left(\frac{1}{e^x}\right) = \frac{1}{1\:+\:x\:+\:x^2/2\:+\:x^3/6\:+\:x^4/24\:+\:x^5/120\:+ ....}$$

Answer 1

This is a bit too large to be a comment.

giancarletto, you might find this to be interesting.

$$ \exp(x)\cdot\exp(-x) $$ $$ =\sum_{n=0}^\infty \frac{x^n}{n!}\cdot\sum_{n=0}^\infty \frac{(-x)^n}{n!} $$ $$ =\sum_{n=0}^\infty \frac{x^n}{n!}\cdot\sum_{n=0}^\infty(-1)^n \frac{x^n}{n!} $$ $$ =\sum_{n=0}^\infty\sum_{i=0}^n (-1)^i \frac1{i!}\frac1{(n-i)!}x^n \quad\mathrm{collecting\ coef\ of\ } x^n $$ $$ = 1+ \sum_{n=1}^\infty \frac{ (1-1)^n}{ n!} x^n = 1 $$ using the binomial theorem on the second to last equality.

Answer 2

Do you know how to rewrite your expression as a power series? You use the geometric series $$\frac 1{1-u} = 1+u+u^2+\dots$$ and substitute, in your case, $u=-(x+x^2/2+x^3/6+\dots)$. For example, if you want to get the terms up the $x^2$ in the expansion, you use the terms up to $x^2$ in $u$ and throw away the higher-order terms. In particular, truncating at $x^2$ in the expression $1-(x+x^2/2) + (x+x^2/2)^2$ gives us $1-x+x^2/2$, as desired.