Why does proving $\lim_{h\to 0} (X(t + h) - X(t)) = 0$ prove continuity?

Question

Why does proving $\lim_{h\to 0} (X(t + h) - X(t)) = 0$ prove continuity?

I don't think it matters, but more specifically, $X(t)$ is a Brownian motion.

I thought that a function $f$ is continuous at the point $x_0$ provided that the image sequence converges to the image point when the sequence $\{x_n\}$ converges to $x_0$.

So wouldn't this only prove continuity at $0$? Or, how does it work?

Answer 1

I thought that a function $f$ is continuous at the point $x_0$ provided that the image sequence converges to the image point when the sequence $\{x_n\}$ converges to $x_0$.

The function is $X$, the sequence is $\{t_h:t_h=(t+h)\}$, the image point is $t_0=t$.

To say that the image sequence converges to the image point is to say $\lim\limits_{h\to 0}X(t_h)=X(t_0)$ or...$$\lim\limits_{h\to 0}\big(X(t+h)-X(t)\big) = 0$$

Thus this is true for all $t$ in the domain, exactly when the function $X$ is continuous everywhere over that domain.

Answer 2

If $t_n \to t$ then $X(t_n)-X(t)=X((t_n-t)+t)-X(t)$ so just note that $h_n=t_n-t \to 0$.