Suppose $f:[3;7]\rightarrow\mathbb{R}$ is continuous, and $f(x)\ge 6$ for all $x\in [3,7)$, prove that $f(7)\ge 6$.
I would just like someone to check my proof.
Suppose to the contrary $f(7)<6$, since $f$ is continuous, we have that there exists $\delta > 0$ s.t. $|x-7|<\delta\implies |f(x) - f(7)|< 6-f(7)$ but then we have $f(x) -f(7) < 6-f(7)$ and so $f(x) < 6$, but since this is only true for $f(7)$, it must be that $\delta=0$ a contradiction.
Is this correct? How could I do it with the sequential definition of continuity. This is not for a class, I am studying because students are asking me these questions in the tutoring lounge I work in and I forgot how to do them...
Using continuity of $f$ we see $$f(7) = \lim_{n\to \infty} f(7-1/n) \geq \lim_{n \to \infty} 6 = 6$$