I think that it makes sense to introduce continuity in the same lesson that you introduce limits.
Here is a sketch of a lesson plan:
Give them this link
https://www.desmos.com/calculator/rlu2zgcjyf
Group work:
- Is $f(2)$ defined or undefined?
- As $t$ approaches $2$, what are the values of $f(t)$ approaching?
- Make a table of values for $t = 1.9, 1.99, 1.999$ and $t=2.1, 2.01, 2.001$.
Summarize and introduce the words and notation for limits. Stress the connection between the numerical and graphical approaches.
Then show them this related graph:
https://www.desmos.com/calculator/zxivpgtrfl
Point out that, in this graph $g(2)$ exists and is equal to the limit which had been calculated before. Define continuity at a point.
Now reveal the formulas:
The formula for the first function is $f(x) = \frac{x-2}{x^2 - 3x + 2}$. Plug in $f(1.99)$ as a sanity check (and to reinforce the numerical/symbolic connection).
Point out that attempting to plug in $x= 2$ results in $\frac{0}{0}$, which is undefined. If we make $x$ close to $2$, but not equal to $2$, then we will get a number close to $0$ divided by a number close to $0$. Spend some time here. If you have clickers you could ask some multiple choice questions like "$\frac{0.0000001}{7}$ is (a) extremely small, (b) extremely large, (c) neither". Then repeat for $\frac{7}{0.0000001}$, $\frac{0.0001}{0.0000001}$, $\frac{0.0000001}{0.0001}$, and $\frac{0.0002}{0.0001}$.
Summarize the findings: when investigating comparing fractions making the numerator smaller makes the fraction smaller, but making the denominator smaller makes the fraction larger. In the case we are looking at there is a "tug of war": both are approaching zero. Which of these drives will win? As the last three numerical examples show in the last paragraph, the answer depends on the relatives sizes of the two quantities, which is hard to predict by just looking at the formula.
Question: can anyone simplify the formula $\frac{x-2}{x^2-3x+2}$?
Answer: Someone gets $g(x) = \frac{1}{x-1}$. Point out that this function is identical to the one we started with, expect it is defined at $x=2$. Plug in $g(1.99)$ as a sanity check. In other words, this is the formula for the second function we graphed.
Can we tell the limit of this function as $x \to 2$ easily? Yes! If $x$ is getting close to $2$, it makes sense that $\frac{1}{x-1}$ is getting close to $\frac{1}{2-1} = 1$.
In other words, we have found $\displaystyle \lim_{x \to 2} f(x)$ by using algebra to find another function $g$ so that $f(x) = g(x)$ when $x \neq 2$, but $g$ is continuous at $x=2$. So finding $\displaystyle \lim_{x \to 2} g(x)$ is easy: just plug it in (direct substitution).
The work I wish I could require for these kinds of problems reflects this way of thinking.
In my ideal world, here is what a student response to the following limit problem would look like:
Find $\displaystyle \lim_{x \to 4} \frac{x-4}{\sqrt{x} - 2}$.
The function $f(x) = \frac{x-4}{\sqrt{x} - 2}$ has a domain of $x \in [0,4) \cup (4, \infty)$. Since $f$ is not defined at $x=4$, it is not continuous at $x=4$.
When $x$ is close to $4$, the formula for $f$ gives me something of the form $\frac{\textrm{close to 0}}{\textrm{close to 0}}$, which is indeterminate. So I will need to do more work to determine the limit.
However,
$$
\begin{align*}
f(x) &= \frac{x-4}{\sqrt{x} - 2}\\
&= \frac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x} - 2}\\
&= \frac{1}{\sqrt{x}+2}
\end{align*}
$$
Thus $g(x) = \frac{1}{\sqrt{x}+2}$ is a new function which is continuous on $[0, \infty)$ and agrees with $f$ everywhere except for $x=4$.
So
$$
\begin{align*}
\lim_{x \to 4} f(x) &= \lim_{x \to 4} g(x)\\
&= g(4)\\
&= \frac{1}{\sqrt{4}+2}\\
& = \frac{1}{2+2}\\
&= \frac{1}{4}
\end{align*}
$$
So $\displaystyle \lim_{x \to 4} \frac{x-4}{\sqrt{x}-2} = \frac{1}{4}$.
I will also note that this kind to analysis pinpoints why so many instructors would be uncomfortable with the work:
$$\displaystyle \lim_{x \to 4} (x+3) = \displaystyle \lim_{x \to 4} 7 = 7$$
even though all of the equalities are correct.