Explanation for an equality (series) $$\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{2} + \frac{x}{{3!}} + \frac{{{x^2}}}{{4!}} + \frac{{{x^3}}}{{5!}} + ...} \right) = \frac{1}{2}$$
Why can one claim this equality?
Thanks.
 A: Note that $e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+...$.
So a bit of algebraic manipulation gives that your limit is same as:
$\lim\limits_{x \rightarrow 0}\dfrac{e^x-1-x}{x^2}$ and you can calculate the limit using L'Hospital's rule if you like.
A: One more approach is based on geometric series. Let $$f(x) = \frac{1}{2} + \frac{x}{3!} + \frac{x^{2}}{4!} + \cdots$$ then we can see that as $x \to 0$ we can assume $|x| < 1$. Now we have $$\begin{aligned}0 \leq \left|f(x) - \frac{1}{2}\right| &\leq \frac{|x|}{3!} + \frac{|x|^{2}}{4!} + \cdots\\
&\leq \frac{|x|}{6} + \frac{|x|^{2}}{18} + \frac{|x|^{3}}{54} + \cdots\\
&= \frac{|x|/6}{1 - (|x|/3)}\text{ (sum of an infinite GP)}\\
&= \frac{|x|}{2(3 - |x|)}\end{aligned}$$ Taking limits as $x \to 0$ we get $$\lim_{x \to 0}\left|f(x) - \frac{1}{2}\right| = 0$$ It follows that that $$\lim_{x \to 0}f(x) = \frac{1}{2}$$
Many times the limit of a power series can be handled using the above technique (which is similar in spirit to the technique given by Did). For a beginner it is better to use such approximation techniques rather than relying on the concept of uniform convergence as applicable to a power series.
A: If you're looking for something which is a little more direct (I suspect that may be the case since this expansion is a common way to calculate the limit given in voldemort's post):
Note $\lim_{x\to 0} \dfrac{x^{n-2}}{n!} = 0\ \forall n> 2$; and $1/2$ for $n=2$ 
Recall that if, for every $n\geq N$:


*

*$\displaystyle\lim_{x\to x_0} f_n(x)$ exists and;

*is equal to $L_n$, say,


Then by additive properties of limits,
$\displaystyle\lim_{x\to x_0} \displaystyle\sum_{n=N}^{\infty} f_n(x)$ exists; and it is equal to $\displaystyle\sum_{n=N}^{\infty} L_n$ (provided that the latter sum converges)
Hence, applying this to $f_n(x) = \dfrac{x^{n-2}}{n!}$ for $n\geq 2$, $x_0 = 0$:
$\displaystyle\lim_{x\to 0} \left( \dfrac{1}{2} + \dfrac{x}{3!} + \dfrac{x^2}{4!} + …\right) = \displaystyle\lim_{x\to 0}  \displaystyle\sum_{n=2}^{\infty} \dfrac{x^{n-2}}{n!} = \dfrac{1}{2} + \displaystyle\sum_{n=3}^{\infty} 0 = \dfrac{1}{2}$
A: For every $|x|\leqslant1$ and every $n\geqslant1$, 
$$-|x|\leqslant x^n\leqslant|x|,$$
hence the sum $f(x)$ of the series considered is such that
$$\frac12-|x|\,\left(\frac1{3!}+\frac1{4!}+\frac1{5!}+\cdots\right)\leqslant f(x)\leqslant\frac12+|x|\,\left(\frac1{3!}+\frac1{4!}+\frac1{5!}+\cdots\right),
$$
in particular,
$$
\left|f(x)-\frac12\right|\leqslant|x|,
$$
which is enough to conclude.
