Gilbert Strang Explanation of Continuous Functions

Question

See snapshot of "Calculus" by Gilbert Strang.

1. What does he mean when he says (sinx)/x can be made continuous at all x. Just set f(0)=1. *Is it not already continuous?

2. The bottom paragraph when he defines a continuable function, I do not understand at all. The only difference between the functions he lists as continuable and not continuable is the continuable ones have points where they shoot off to infinity. But he doesn't say that. He says " all x in a way that makes it continuous". How can he say that sqrt(x) continuity can be extended to all x when it isn't defined below zero.

Answer 1

The function $(\sin x)/x$, as it is, is not defined at $x=0$ since you cannot divide by zero. What he means is that you can define a new function $$ f(x)=\begin{cases} \frac{\sin x}{x} & \text{if $x\ne 0$} \\ 1 & \text{if $x=0$} \end{cases} $$ which is continuous at $x=0$. So, in some sense, we made $(\sin x)/x$ continuous at 0 and, in turn, at all $x$.

The reason you can do that, and to answer your second question, is that, for continuous functions, $$ \lim_{x\to x_0} f(x) = f(x_0) $$ where $f(x_0)$ is finite. It is a good exercise to derive this fact from the $\epsilon$ / $\delta$ definitions of continuity at $x_0$ and limit as $x\to x_0$.

In our case, as we said before, $(\sin x)/x$ is not defined at $x=0$ because of division by 0, but we do have $$ \lim_{x\to 0} \frac{\sin x}{x} = 1 $$ and, thus, we can extend $(\sin x)/x$ continuously (make it continuous) at 0.

Conversely, $1/x$ is also not defined at $x=0$ but, this time, $$ \lim_{x\to 0^-} \frac{1}{x} = -\infty $$ and $$ \lim_{x\to 0^+} \frac{1}{x} = +\infty $$ which means we cannot find a "continuous version" of $1/x$. A similar issue happens with $\tan x$ at $x=\pi/2 + k\pi$, $k\in\mathbb{Z}$. As for $\sqrt{x}$, my guess is that he had defined it only for positive $x$ and then he extends it continuously at $x=0$, as I agree with you that we can't take $x <0$ (at least, not without bringing complex numbers into the game).

Answer 2

1. f is discontinuous at 0 because it is undefined at 0 -- look up the definition of continuity as a limit or epsilon-delta argument. On the other hand, since the limit of f exists at 0 (it's 1), simply redefining the function there as its limit will make the function continuous on all real numbers. Such discontinuities are called holes or removable discontinuities.

2. Strang could have been more explicit. For example, the Dirichlet function [https://mathworld.wolfram.com/DirichletFunction.html] is discontinuous everywhere as is its restriction to the irrational numbers union {m/2^k: m, k are integers with k > 0} -- yet defining/extending this function on the rationals (a null-set or a set with zero Lebesgue measure) to have the same value as on the irrational reals makes this simply a constant function, which is continuous for all real numbers.