Why can we use Taylor series for evaluating limits? [duplicate]

I was trying to solve a limit with Taylor series today, which I never had a problem with, but now I suddenly don't understand why it's ok to do it in that way.

For example, the limit

$$\begin{align} &\lim_{x \to 0} \frac{x - x^3/6 - \sin{x}}{x^5} \\ &= \lim_{x \to 0} \frac{(x-x^3/6) - (x - x^3/6 + x^5/5! - \text{higher order terms} )}{x^5} \\ &= \lim_{x \to 0} \frac{-x^5/5! + \text{higher order terms}}{x^5} \\ &= \lim_{x \to 0} {\frac{-1}{5!} + \text{higher order terms}} = - \frac{1}{120} \end{align}$$

But why can we say that the sum of higher order terms goes to zero? I understand that each of the terms goes to zero, but why does their infinite sum go to zero? How can I justify that? I think that although the summands are small, their infinite sum can hypothetically still be big?

I would write it $$\begin{align} &\lim_{x \to 0} \frac{x - x^3/6 - \sin{x}}{x^5} \\ &= \lim_{x \to 0} \frac{(x-x^3/6) - (x - x^3/6 + x^5/5! +O(x^6) )}{x^5} \\ &= \lim_{x \to 0} \frac{-x^5/5! + O(x^6)}{x^5} \\ &= \lim_{x \to 0} \left({\frac{-1}{5!} + O(x)}\right) = - \frac{1}{120} \end{align}$$ and say that the first equality is due to "Taylor's theorem".

We can do this by Taylor's Mean Value Theorem .

For a $k$ times differentiable function at a point $a$.

$f(x)=f(a)+(x-a)\frac{f^(1)(a)}{1!}+(x-a)^{2}\frac{f^{(2)}(a)}{2!}+...+(x-a)^{k}M_{k}$

Where $M_{k}\to 0$ as $x\to a$.