Is this function continuous at x=2?

Question

I have a function, k, below that I think is not continuous at x=2, but I'm not sure. If it is how can it be proven.

Let: $h(x) = \frac{x^2+20}{6} $

$g(x) =\frac{12+8x-x^2}{6}$

$t(x) =4+\frac{2}{3}(x-2)$

Notice: g(x)≤h(x), g(2)=h(2)=4 and g'(2)=h'(2)=2/3

(plot of the 3 functions)

Now Let:

I think that function k, is NOT continuous because there is an irrational number between any two unequal rational numbers and between any two distinct irrational numbers, there exists a rational number. Both pieces of function k will converge at the point (2,4) but I don't see how that will make it continuous.

Please help me clarify this.

Answer 1

The function is indeed continuous. That is because continuity at $x_0$ is defined as $\lim \limits_{x \to x_0} h(x) = h(x_0) $

Now consider the limit for $ x \in \mathbb{Q} $. $\lim \limits_{x \to 2} h(x) = 4 $ Similarly for $ x \in \mathbb{Q^c}$

So $ \forall ε > 0$ $\exists δ_1, δ_2 > 0$ such that $\forall x \in \mathbb{Q}$ where $ |x - 2| < δ_1 $ and $\forall x \in \mathbb{Q^c}$ where $ | x - 2| < δ_2 $ $|k(x) - 2|< ε$

Then just take the minimum of $δ_1, δ_2$