Intuitive motivation for limit computations This is a Q&A pair concerning intuitive motivation for limit computations. Usually, my standard advice is to use asymptotic expansions to compute limits (especially for harder things like this or this), but if we wish to do it without asymptotic expansions yet in a well-motivated way, we may want to have some intuitive explanation for why various elementary tricks work.
For example, to prove that $\dfrac{1+2x-\sqrt[3]{1+6x}}{x^2} ≈ 4$ as $x → 0$, an elegant way is to let $p = 1+2x$ and $r = \sqrt[3]{1+6x}$, so as $x → 0$ we have $p,r → 1$ and hence $\dfrac{p-r}{x^2}$ $= \dfrac{p^3-r^3}{x^2·(p^2+p·r+r^2)}$ $= \dfrac{12+{?}x}{p^2+p·r+r^2}$ $≈ \dfrac{12}{1+1+1}$.
This trick may seem mysterious. After all, why did 'multiplying by the conjugate' work, and it is always possible to find such tricks? What if we are asked to find $\lim_{x→0} \dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}$? Is there a systematic yet intuitive way to figure out that we can apply the above trick to both parts? Personally, I prefer computing it by asymptotic expansion, namely that as $x → 0$ we clearly have $\sqrt{1+4x} ∈ 1+2x-2x^2+o(x^2)$ and $\sqrt[3]{1+6x} ∈ 1+2x-4x^2+o(x^2)$, and so the result follows quickly.
But the question remains: Can we find the asymptotic expansion intuitively without higher-power tools (such as Taylor series or binomial expansion for non-natural powers)? And better still, can we find an elementary solution without even rigorously proving the asymptotic expansion?
 A: Here is one way to do it. The idea is to non-rigorously obtain successively better approximations, and then turn those into rigorous elementary limit computations.
For the example in the question, we wish to approximate $\sqrt{1+4x}$ and $\sqrt[3]{1+6x}$ as $x → 0$. Clearly, the first-order approximation for each of them is $1$, but that is clearly insufficient because we have no bound on their difference as $x → 0$. So we want a better approximation.
Since $(1+2x)^2$ matches the "$4x$" term, we guess that $a = 1+2x$ better approximates $\sqrt{1+4x}$. How much better? $a^2 = (1+4x) + 4x^2$, so the 'remainder' is $4x^2$ (compared to the 'remainder' $4x$ for approximation $1$). If we change the coefficient of $x$ to anything else, the 'remainder' will have a nonzero '$x$ term' which is much bigger than $4x^2$ as $x → 0$, so we intuitively know that $2$ is really the best coefficient.
Doing the same for $\sqrt[3]{1+6x}$, we intuitively get $1+2x$ as the best approximation up to the '$x$ term'. Again, we can intuitively see that these approximations are too poor since they cancel. So we try adding a suitable '$x^2$ term'. For $\sqrt{1+4x}$, we want $(1+2x+cx^2)^2$ to match $1+4x$ up to as high power of $x$ as possible. Note that what we have is $(a+cx^2)^2$ where $cx^2$ will be negligible compared to $a = 1+2x$, so the main term of $(a+cx^2)^2 - a^2$ is $2acx^2$. Recall that using $a^2$ to approximate $1+4x$ left a 'remainder' of $4x^2$, which we can cancel by setting $c = -2$. Thus we intuitively know that $1+2x-2x^2$ is the best approximation for $\sqrt{1+4x}$ up to the '$x^2$ term'.
By similar reasoning, we can find that $1+2x-4x^2$ is the best approximation for $\sqrt[3]{1+6x}$ up to the '$x^2$ term' and we can intuitively see that this is enough to guess that $\dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}$ $≈ \dfrac{(1+2x-2x^2)+(1+2x-4x^2)}{x^2} = 2$ as $x → 0$, since the 'remainder' terms that we neglected are negligible compared to $x^2$.
Note that all the above reasoning is not rigorous, but once we find these approximations we can readily apply the 'conjugate trick'. Specifically let $p = 1+2x$ and $q = \sqrt{1+4x}$ and $r = \sqrt[3]{1+6x}$, so as $x → 0$ we have $p,q,r → 1$ and hence:
  $\dfrac{q-r}{x^2}$ $= \dfrac{q-p}{x^2} + \dfrac{p-r}{x^2}$
  $= \dfrac{q^2-p^2}{x^2·(q+p)} + \dfrac{p^3-r^3}{x^2·(p^2+p·r+r^2)}$
  $= \dfrac{-4}{q+p} + \dfrac{12+{?}x}{p^2+p·r+r^2}$
  $→ \dfrac{-4}{1+1} + \dfrac{12}{1+1+1}$.
The only remaining mystery is whether this 'conjugate trick' is just a trick that works for surds and nothing else. Perhaps not surprisingly, it can be generalized. In general it can be used whenever we want an approximation for $f^{-1}(f(c)+t)$ as $t → 0$, where $f$ is has nonzero derivative at $c$ and is invertible in an open interval around $c$.
Specifically, $\dfrac{f^{-1}(f(c)+t)-c}{t}$ $≈ (f^{-1})'(f(c))$ $= \dfrac1{f'(c)}$ as $t → 0$. For the above example, we would let $f(x) := x^3$ for each real $x$, and so $\dfrac{\sqrt[3]{p^3-s}-p}{s} ≈ \dfrac1{3p^2}$ where $s = 12x^2+{?}x^3 → 0$ as $x → 0$.
So effectively the trick is simply an unfolding of the proof of the derivative of $(f^{-1})'$ at $f(c)$. Hence if you have an elementary proof of that, you can use it to obtain a rigorous elementary proof of the associated limit.
I hope that this explanation shows how one can use non-rigorous intuition to derive a rigorous proof of certain kinds of limits without using high-power tools. We do need some real analysis to truly understand why this method of finding such proofs works, but only the most rudimentary limit concepts are actually used by the proofs produced by the method. So one can teach this method as a rigorous tool prior to teaching real analysis, and yet strongly motivate the intuitive concept of asymptotic expansion.
