Given an infinitely small quantity: $$\alpha \left ( x \right )= \tan \left ( x \right )-\sin \left( x \right)$$ as x aproaches $0$, and computing the corresponding asymptotic relationship. What does all of that mean?
By definition, a function $f$ is to be called infinitely small with respect to the function $g$ as $x\rightarrow a$ and thus we write $f\left ( x \right )=o\left ( g\left ( x \right ) \right )$ , $x\rightarrow a$, if there exists such a neighborhood of $a$, that $f\left ( x \right )=\alpha \left ( x \right )g\left ( x \right )$ for every $x\in U_{D}$, where $\alpha(x)$ is an infinitely small function as $x$ approaches $a$.
I can't seem to find the link between the given infinitely small quantity and the one in the definition. My initial guess was to find the series expansion of $\alpha$ about $0$. Hence I got $$\alpha \left ( x \right )= \frac{x^{3}}{3}+o\left ( x^{4} \right )$$ How do I know of which order this is, and what does it mean to find the asymptotic relationship for this? I know I could just find the answer in the solution-sheet , but what is the idea behind this, and how do I grasp it intuitively?
I think (I'm not expert here at all!) that you want to say that $f$ is infinitely small with respect to some polynomial (given your definition) like $(x-a)^n$.
For instance, I think that as you approach 0, the function $$f(x) = x$$ is infinitely small compared to $$g(x) = 1,$$ but not compared to $$g(x) = x,$$ etc.
If $f$ is infinitely small wrt $$g(x) =(x-a)^n,$$ we say it has order at least $n+1$ (at $a$). If it's also not infinitely small wrt $g(x) = (x-a)^{n+1}$, we say it has order exactly $n+1$ at $a$.
In your case, your Taylor series seems to show that your function is small to order 3.