Error evaluating $ \lim_{x\to 0}\frac{x-\tan x}{x^3} $

The explanation you're looking for is this. You are implicitly using the properties $\lim\limits (f+g)=\lim\limits(f)+\lim\limits(g)$ and $\lim\limits (fg)=\lim\limits(f)\lim\limits(g)$, but this is only true when all the limits in these equalities exist, (disclaimer: there are important assumptions that I'm not writing but that should accompany these properties). More specifically, what you did (implicitly) was: $$ \begin{align} \lim\limits_{x\to 0} \left({1\over x^2} - \frac{\tan x}{x^3}\right) &=\lim\limits_{x\to 0}\left({1\over x^2} - {\tan x\over x}\cdot {1\over x^2}\right)\\ &=\lim\limits_{x\to 0}\left({1\over x^2}\right) + \lim\limits_{x\to 0}\left(- {\tan x\over x}\cdot {1\over x^2}\right) \tag{Incorrect}\\ &=\lim\limits_{x\to 0}\left({1\over x^2}\right) - \lim\limits_{x\to 0}\left( {\tan x\over x}\right)\lim\limits_{x\to 0}\left({1\over x^2}\right) \tag{Incorrect}\\ &=\lim\limits_{x\to 0}\left({1\over x^2}\right) - \lim\limits_{x\to 0}\left({1\over x^2}\right) \tag{*}\\ &=\lim\limits_{x\to 0}\left({1\over x^2} - {1\over x^2}\right) \tag{Incorrect}\\ &=0 \tag{**} \end{align} $$

$(\text*)\text{ As correct as something meaningless can be}$
$(\text{**})\text{ Actually correct, but it's too late}$

You can't just replace the value of the limit inside without using the above reasoning or something else which will end up not working.

Although, in fact, $\lim_{x\to0}\frac{\tan x}x=1$, you cannot deduce from that that$$\lim_{x\to0}\frac1{x^2}-\frac{\tan x}x\times\frac1{x^2}=\lim_{x\to0}\frac1{x_2}-\frac1{x^2}.$$

In this case, L'Hopital's Rule is the way to go:\begin{align}\lim_{x\to0}\frac{x-\tan x}{x^3}&=\lim_{x\to0}\frac{-\tan^2x}{3x^2}\\&=-\frac13\left(\lim_{x\to0}\frac{\tan x}x\right)^2\\&=-\frac13.\end{align}

Right, $$\lim_{x\to0}\frac{\tan x}x=1.$$

But that doesn't mean that you can replace $\dfrac{\tan x}x$ by $1$ inside the limit !

Actually,

$$\frac{\tan x}x=1+f(x)\ne1$$ and the function $f$ can strike back.

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The "striking back" works like this:

• subtracting $1$ from $\dfrac{\tan x}x$ isolates $f(x)$.

• then dividing by $x^2$ "amplifies" it, giving the term $\dfrac{f(x)}{x^2}$. It turns out that $f(x)$ has a double root at $x=0$, so that the fraction bring a finite contribution, but it could very well have been unbounded.