Is $f(x)=\lim_{m\to\infty}m(x-b)$ the same as $x=b$?

Question

I was thinking that maybe a linear function with an infinite slope $$f(x)=\lim_{m\to\infty}m(x-b)$$ would be the same as $x=b$. And I saw another question Is the slope of a vertical line infinity or undefined? where they say the slope of a vertical line is undefined, so I'm not sure on this.

Answer 1

As mentioned in the comment (by Randall), the expression $$ f(x) = \lim_{m\to \infty} m(x-b) $$ that you wrote doesn't make sense as a real-valued function. Even if we think of $f$ as an extended real-valued function $f:\Bbb R \to [-\infty,\infty]$, we still have $$ f(x) = \begin{cases} -\infty &; x<b \\ 0 &; x=b \\ \infty &; x>b, \end{cases} $$ which is probably not what you are looking for.

However, there is a way to make sense of what it means for the sequence of lines $$L_m = \{ (x,y) \in \Bbb R^2 : y = m(x-b)\}$$ to converge to the line $$L_{\infty} = \{ (x,y) \in \Bbb R^2 : x=b \}$$ as $m\to\infty$. One way to do that is to use the concept of Kuratowski convergence, which makes precise how a sequence of sets can be said to converge to a limiting set. In this setting, we can rigorously say that the graphs of the functions $$ f_m(x) = m(x-b) $$ converge (in Kuratowski's sense) to the line $x=b$ in $\Bbb R^2$ (since $\text{Gr}\, f_m = L_m$). That being said, the limiting object $L_\infty$ does not correspond to a graph of any real-valued function on $\Bbb R$, and it can be misleading to think of it as a "line with infinite slope".