<s>[INST] Limit: $\lim_{x\to 0}\frac{x^3 - 3x^2+1}{x^3}$ I want find the limit of $\frac{x^3 - 3x^2+1}{x^3}$ as $x\to 0$. 
So I write
$$\frac{x^3 - 3x^2+1}{x^3} = 1 - \frac{3x^2 +1}{x^3}$$ and
$$\lim_{x \to 0^- }(\frac{x^3 - 3x^2+1}{x^3})= \lim_{x \to 0^-}(1 - \frac{3x^2 +1}{x^3})$$ Here, the limit must be  $\pm \infty$ and I think when $x \to 0^-$ as $x^2$ is something positive, $x^3$ is something negative and since we have $- \frac{3x^2 +1}{x^3}$ we should have $+\infty$. However, it must be  $-\infty$. Similarly I think $\displaystyle\lim_{x \to 0^-}(1 - \frac{3x^2 +1}{x^3}) =-\infty$ but it's $+ \infty$.
What do I miss, how should I think?
 [/INST] You made a sign mistake,  $$\frac{x^3 - 3x^2+1}{x^3} = 1 - \frac{3x^2 \color{red}{-}1}{x^3}$$
Hence your conclusion is opposite.
Intuitively, as  $x \to 0^-$,  $x^2$ is small in magnitude and hence  $(3x^2-1)$ approaches $-1$, $x^3$ is something negative and it is small, so we have $- \frac{3x^2 +1}{x^3}$ goes to $-\infty$. 
The left hand side limit goes to $-\infty$ while the right hand side limit goes to $+\infty$, hence the limit doesn't exist.
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