Error evaluating limx→0x−tan⁡xx3 I solved it like this, limx→0(1x2−tan⁡xx3)=limx→0(1x2−tan⁡xx⋅1x2) Now using the property limx→0tan⁡xx=1 we...

The explanation you're looking for is this. You are implicitly using the properties ${\displaystyle lim(f+g)=lim\left(f\right)+lim\left(g\right)}$ and ${\displaystyle lim\left(fg\right)=lim\left(f\right)lim\left(g\right)}$, 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:
${\displaystyle \underset{x\to 0}{lim}(\frac{1}{x^2}-\frac{\tan x}{x^3})=\underset{x\to 0}{lim}(\frac{1}{x^2}-\frac{\tan x}{x}\cdot \frac{1}{x^2})}$
${\displaystyle =\underset{x\to 0}{lim}\left(\frac{1}{x^2}\right)+\underset{x\to 0}{lim}(-\frac{\tan x}{x}\cdot \frac{1}{x^2})}$ (Incorrect)
$=\underset{x\to 0}{lim}(\frac{1}{x^2})-\underset{x\to 0}{lim}(\frac{\tan x}{x})\underset{x\to 0}{lim}(\frac{1}{x^2})$ (Incorrect)

$=\underset{x\to 0}{lim}(\frac{1}{x^2})-\underset{x\to 0}{lim}(\frac{1}{x^2})(\ast )$
$=\underset{x\to 0}{lim}(\frac{1}{x^2}-\frac{1}{x^2})$ (Incorrect)
$=0$ (**)
(*) As correct as something meaningless can be
(**) Actually correct, but it's too late