I've been reviewing some simple calculus style material but I think my knowledge is incomplete.

Suppose we are trying to calculate $\displaystyle\lim_{x \rightarrow 0} \frac{f(x)}{g(x)}$, where $f,g$ for our purposes are smooth.

A useful approach is to Taylor expand the numerator and denominator in the hope of removing any interfering terms that prevent directly evaluating the limit, example here: Using Taylor Expansion to evaluate limits

What I don't understand about this method is to what order should we expand to? In the example above, the third order expansion of $\sinh(x)$ is taken giving a $O(x^{13})$ term and a second order expansion of $\sin(x)$ is taken to get another $O(x^{13})$ term (after some truncation of higher order terms from the binomial expansion, I think?)

Does this mean I am aiming for something like $\displaystyle\lim_{x \rightarrow 0} \frac{r(x) + O(x^p)}{q(x) + O(x^p)}$ for polynomials $r,q$ and $p \in \mathbb{N}$ is allowed to be anything I want provided I can "match" the asymptotic terms?

In the case of $\displaystyle\lim_{x \rightarrow c}$ the procedure is the same but with Taylor expansions centered around $c$? Out of curiousity can this be extended somehow for$\displaystyle\lim_{x \rightarrow \pm \infty}$?

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    $\begingroup$ You can use $\lim_{x\to\pm\infty}\phi(x)=\lim_{z\to0}\phi\left(\frac1z\right)$. $\endgroup$ – Yves Daoust Sep 11 '18 at 18:06

The goal of series expansion is to find the term of the lowest degree in the development (smooth functions behave like a polynomial of infinite degree near zero), because it dominates the other terms for a limit to zero.

For an indeterminate form of the type $\dfrac00$, we establish

$$\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{ax^n+o(x^n)}{bx^m+o(x^m)}=\lim_{x\to0}\frac abx^{n-m}$$ and the final result depends on the sign of $n-m$.

If you expand with too many terms, this is harmless. For instance,

$$\lim_{x\to0}\frac{ax^n+bx^{n+1}+o(x^{n+1})}{cx^m+dx^{m+1}+o(x^{m+1})}=\lim_{x\to0}\frac acx^{n-m}$$ as well.

But you must make sure to expand enough terms when there is cancellation (subtraction of asymptotically similar functions), because some of them may vanish.


$$f(x)=\sin x-x=x+o(x)-x=o(x)$$ doesn't work because there is no leading term. With one more,

$$f(x)=\sin x-x=x-\frac{x^3}6+o(x^3)-x=-\frac{x^3}6+o(x^3).$$

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    $\begingroup$ Thanks for the fast and very comprehensive answer, just what I needed. $\endgroup$ – Theo Diamantakis Sep 11 '18 at 18:11

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