Why can the Taylor formula be used to evaluate the limit,Why does its remainder converge to zero？ A simple example $\lim\limits_{x\to0}\frac{\cos(x)-1}{x^2}=\frac{1-\frac{1}{2}x^2+o(x^2)-1}{x^2}=-\frac{1}{2}$

If taken out alone, its limit is zero.How can I judge that the sum of infinite limits converges to zero.
Such as:$\lim\limits_{x\to0}\frac{o(x^2)}{x^2}=\lim\limits_{n\to\infty}[\sum\limits_{i=3}^{n} (\lim\limits_{x\to0}\frac{x^i}{x^2})]=?$
Clarification of the problem
For example $\lim\limits_{n\to\infty}(\frac{1}{n}+\frac{1}{n}+.....+\frac{1}{n})=\lim\limits_{n\to\infty} n*\frac{1}{n}=1$
In the case of an infinite number of items,I cannot simply think of adding an infinite number of zero
 A: Apply the rules of asymptotic analysis:
$$\frac{\cos x-1}{x^2}=\frac{1-\frac{1}{2}x^2+o(x^2)-1}{x^2}=-\frac 12+o(1).$$
Added: an elementary way to determine the limit:
$$\frac{\cos x-1}{x^2}=\frac{\cos^2 x-1}{x^2(\cos x +1)}=-\underbrace{\frac{\sin^2x}{x^2}}_{\substack{\downarrow\\1^2=1}}\,\underbrace{\frac1{\cos x+1}}_{\substack{\downarrow\\1+1=2}}$$
A: It really depends on how exact you need it. You could use the remainder term of the Taylor expansion of degree $n$ of some $f$ around some $x_0$ in its more detailed form, $$r_n(x)=\frac{(x-x_0)^{n+1}}{(n+1)!}f^{(n+1)}(\xi),~~~\xi\in[x_0,x].$$
For the cosine with $x_0=0$ this would give for example
$$
r_3(x)=\frac{x^4}{4!}\cos(\xi)
$$
Or you could refer back to more elementary means and notice that the cosine series is alternating with falling terms if $x$ is small enough, so that the series is bounded by its partial sums,
$$
1-\frac{x^2}2\le\cos(x)\le 1-\frac{x^2}2+\frac{x^4}{4!}
$$
All this is showing that the defect in the limit formula is smaller than $\frac{x^2}{24}$ which does indeed converge to zero.
