proving piece wise function is discontinuous

Question

Visually, we can check to see that the following function is discontinuous $$ f(x)=\begin{cases} 3x^2+\cos(2\pi x) & x\lt 3\\ x-3 & x\geq 3 \end{cases} $$

But how do we prove it?

Answer 1

• $\lim_{x\to3^+} f(x) = \lim_{x\to3^+} (x - 3) = 3 - 3 = 0$
• $\lim_{x\to3^-} f(x) = \lim_{x\to3^-} (3x^2+\cos(2\pi x)) = 3(3^2)+\cos(6\pi) = 27+1 = 28 \ne 0$

Hence $\lim_{x\to3^+} f(x) = f(3) \ne \lim_{x\to3^-} f(x)$. $f$ is discontinuous at $x=3$.

Answer 2

we have $$\lim_{x\to 3^-}(3x^2+\cos(2\pi x))=27+\cos(6\pi)=...?$$