Calculating $\lim_{x\to0^+}x-\frac{1}{x^3}$

Question

$$\lim_{x\to0^+}x-\frac{1}{x^3}$$

The answer is $-\infty$.

It's not very clear to me how was that concluded. You can't plug in the $0$ because you'd have $\frac{1}{0}$ which is indeterminate. But then again, I don't see much room for algebraic trickery to change the shape of the limit.

How can I proceed here?

Answer 1

Let $M < 0$; if $x > 0$ such that $x < \frac{1}{2}$ and $< \big( \frac{1 - 2^{-4}}{|M|} \big)^{1/3}$, then $$ x - \frac{1}{x^{3}} < \frac{1}{2} - \frac{1}{x^{3}} = \frac{x^{3} - 2}{2x^{3}} < \frac{2^{-3}-2}{2x^{3}} = \frac{2^{-4} -1}{x^{3}} < M. $$

Answer 2

Substitute $y=\frac1x$. So as $x\to0^+, y\to\infty$. Thus our limit becomes $$\lim\limits_{y\to\infty}\frac1y-y^3=0-\infty=-\infty$$
We know that $\lim\limits_{x\to0}=\frac1x=[\text{Undefined}]$, but if we calculate the RHL, we will get $\infty$. (LHL is $-\infty$)
And $(\infty)^3=\infty$