$\sqrt{3} - 0m = 0$ only if $m = \infty$ … What?!

On "Introduction to Discrete Dynamical Systems and Chaos" by Mario Martelli, at 86 pp, it's stated the implication used as title for this question: $$"\ldots\sqrt{3} - 0m = 0 \text{ , which can be true only if } m=\infty."$$ but I don't understand the algebra of this reasoning. The expression is equivalent to $$\sqrt{3} = 0$$, that should not be always false on the Real line? Introducing $$0m$$ and setting $$m=\infty$$ isn't rising an indeterminate form $$0\infty$$?

• It's a sloppy way of saying $0m=0$ whenever $m$ is finite. But interpreting $0 \cdot \infty$ as a finite nonzero number really requires embedding it in some surrounding context, such as a limit of $f(x) g(x)$ where $f(x) \to 0$ and $g(x) \to \infty$. – Ian Oct 10 at 14:18
• Yes, $0\infty$ is indeterminate, so it can take any value, in particular $\sqrt 3$. Of course, $-\infty$ can also be used. Anyway, I assume there is more context to it than you present in this question. Is $m=\infty$ important? – Andrei Oct 10 at 14:19
• Yes @Andrei, $m$ is an angular coefficient used in the state-space analysis. – Informatico di campagna Oct 10 at 14:22
The question discussed in the book is about the tangent lines of $$0=G(x,y)=x^2-xy+y^2-1$$ in several points. First $$(x_0,y_0)=(1,1)$$ is discussed for the regular case, then the special situation in the point $$(x_0,y_0)=(2/\sqrt3,1/\sqrt3)$$ is explored. As the equation of the tangent line in general is $$0=(2x_0-y_0)(x-x_0)+(2y_0-x_0)(y-y_0),$$ one sees that this tangent is vertical because of $$2y_0-x_0=0$$. That the slope $$m=dy/dx$$ is $$∞$$, that is, no finite slope exists, just means that the tangent can not be written as a function of $$x$$.