calculus continuity function of unknown

Question

I am watching a video on Introduction of Continuity, and I am confused by this statement made by the professor: "when $x$ is lesser than $1$, it wants to be $ax - 4$, but I want, when $x$ is $1$, it to be $a\cdot 1 - 4 = 3$, which is equal to $a - 4 = 3$, which implies $a$ is $7$."

Since $x$ is $1$, is satisfies the condition $x \geq 1$, and it should evaluate to $3(1)^2$ instead of $a(1)-4=3$.

Can anyone explain?

Answer 1

The problem we're trying to solve is:

Find all values of $a$ such that $f$ is continuous everywhere.

In particular, we need $f$ to be continuous at $x = 1$. By definition, this means that:

$$ \lim_{x \to 1} f(x) = f(1) = 3 \cdot 1^2 = 3 $$

For the two-sided limit to equal some number (in this case, $3$), we at least need the two-sided limit to exist. To guarantee existence, each of the one-sided limits must also exist and match each other. So in particular, we know that:

$$ \lim_{x \to 1^-} f(x) = 3 $$

But if $x$ is approaching $1$ from the left, then $x < 1$, so we know which piece of $f$ to use. This means that:

$$ \lim_{x \to 1^-} (ax - 4) = 3 \iff a \cdot 1 - 4 = 3 \iff a = 7 $$