How to justify this general Taylor series argument given by physicists?

Question

I'm a 3rd year Physics and Mathematics student, an in this 3 year, I always faced with arguments like this;

Let $f(t)$ be a function of $t$. Then, for small $\delta >0$, we can expand $f$ around $0$ by its Taylor series, so

$f(\delta) = f(0) + \delta \frac{d f}{dt}|_{t = 0} + O(\delta^2)$.

Now, since $f$ satisfies this $\dots$ equality, by putting its taylor expansion in that equality, we get $$\text{(some terms involving $\delta$)} + O(\delta^2) = 0.$$ Since, the terms in $O(\delta^2)$ are much small compared to the other terms, in the limit $\delta \to 0$ we get

$$\text{(some terms involving $\delta$)} = 0,$$ so this means the coefficient of $\delta $ in LHS has to be zero, hence $\dots$.

I know that while doing physics, we do lots of assumption without explicitly stating them; however, since we are just building a model, as long as I can build a consistent machinery to duplicate the principles of the phenomena at hand, there is no problem.

Question:

However, assuming that the given above "mathematical" argument is still in the bounds of the known mathematics to mankind, how can one justify that, indeed, the coefficients of $\delta$ in the last line should be zero ?

To give a little more concrete example, say $f(0) = 0$ and the second term in the expansion is $\delta (K - K')$, so by the same argument, $$\delta (K- K') + O(\delta^2) = 0 \Rightarrow K=K'.$$

But, in such a case, if $\delta \to 0$, then the term $\delta (K-K') \to 0$ or equal to zero, but a such a general argument ignores the fact that $\delta (K-K')$ might not be equal to zero, but is quite small, in which case $K = K'$ is not necessarily true.

Edit:

An example of such argument given in https://www.youtube.com/watch?v=3FTyQS3y2UU&feature=youtu.be&t=2466.

Answer 1

I don't see your point. The expression: $$ \delta(K−K′)+O(\delta^2)=0 $$ means that the LHS is exactly $0$, not only in the limit $\delta\to0$ but for all values of $\delta$ in a certain neighbourhood of $0$ (at least). Dividing then both sides by $\delta$ and taking the limit $\delta\to0$ gives $K-K'=0$. That's perfectly rigorous.

Answer 2

For a function differentiable in a neighbrohood of $0$, we can write rigorously

$$f(\delta)=f(0)+f'(0)\delta+o(\delta).$$

So if by other means you establish that

$$f(\delta)=f(0)+o(\delta^2),$$ then certainly $f'(0)=0$.