Note in general $x \neq \sqrt{x^2}$ because RHS is always positive, whereas LHS can be negative if $x<0$ (for instance $-1\neq \sqrt{(-1)^2}=1$). So in the second step if $x\to -\infty$ then clearly $x^3<0$, so we have
$$
1/x^3=-\sqrt{1/x^6}
$$
Because from my understanding, in order for it to be a tangent line, it intersects the curve at one point only, however Δx approaches zero, it never reaches it, so Δx must be greater than zero, however infinitesimally small, correct?
You're right. We don't ever reach that point. We take a limit.
The colloquialism, "reaching the point" is a good anthropomorphic description. Limits allow us to stretch the constraints of the real numbers by pushing towards the infinite and infinitesimal. Technically, though, to venture into such territory, we need to properly define limits. This is often introduced with the epsilon-delta formalization.
Say there exists a limit $f'(x)=\lim_{\Delta x\rightarrow0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$. Then for every $\epsilon>0$, there exists some $\delta>0$ such that whenever $0<\Delta x<\delta$, we find $|f'(x) - \frac{f(x+\Delta x)-f(x)}{\Delta x}|<\epsilon$.
We can heuristically think of the last paragraph as the following: our derivative exists if for every positive number $\epsilon$ and $\delta$, including the most ridiculously small numbers you can ever imagine, whenever $\Delta x$ is trapped between zero and any of these ridiculously small numbers, the difference between our derivative and the original expression is imperceptible.
But wait a minute, you say
...Δx must be greater than zero, however infinitesimally small, correct?
The epsilon-delta definition seems to hint that as well, but there's a catch: $$|f'(x) - \frac{f(x+\Delta x)-f(x)}{\Delta x}|<\epsilon$$
This is not less than some real positive number $\epsilon$. This is less than ANY POSSIBLE real positive number $\epsilon$. Such a concept only exists within the formalism of a limit, and is by no means a measurable quantity. That's what is meant by infinitesimal.
Due to the limit, then, the derivative cannot represent any possible secant line. There are no two points corresponding to $x+\Delta x$ and $x$ that are indistinguishable! The value we reach has converged to that which represents the slope of the tangent.
Added note:
$\Delta x\rightarrow 0$ doesn't just imply that $\Delta x$ is running through the positive numbers towards zero. For the limit to exist, we typically require it to be two-sided, meaning that $\Delta x\rightarrow0^+$ and $\Delta x\rightarrow0^-$ must produce the same result. In either case, the difference between $\Delta x$ and zero becomes vanishingly small.
Best Answer
$h$ is not "one point on the secant line", it is the horizontal distance between the two points on the secant line. So saying "$h$ goes to $0$" means "Let the two points close in on eachother".