Limit: $\lim_{x\to 0}\frac{x^3 - 3x^2+1}{x^3}$

Question

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?

Answer 1

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.

Answer 2

Write $$f(x)=g(x)h(x),$$ where $$g(x)=x^3-3x^2+1,\quad h(x)=\frac{1}{x^3}.$$ Now, approaching from the right, we have $$\lim_{x\to 0^{+}}g(x)=1,\quad \lim_{x\to 0^{+}}h(x)=\infty.$$ Similarly, approaching from the left, we have $$\lim_{x\to 0^{-}}g(x)=1,\quad\lim_{x\to 0^-}h(x)=-\infty.$$ The limits regarding $h(x)$ can be seen by looking at the sketch for the corresponding graph.

Since the right hand/left hand limits aren't equal; $\lim_{x\to 0}f(x)$ does not exist.