[Math] Proving discontinuity of a piecewise function

Question

Let
$$
f(x) = \left\{
\begin{array}{ll}
x^2 & \quad 0 \leq x < 1 \\
2 & \quad 1 \leq x
\end{array}
\right.
$$
-$f(x)$ is discontinuous at $x=1$ as it is a jump discontinuity.

-Let $$a_n = 1 + \frac{1}{n}$$

$$b_n = 1 – \frac{1}{n}$$

-For both of these sequences, as $$n \rightarrow \infty, a_n,b_n \rightarrow 1$$-It then follows that:

$$
f(a_n) = \left\{
\begin{array}{ll}
(1 + \frac{1}{n})^2 & \quad 0 \leq n < 1 \\
2 & \quad 1 \leq n
\end{array}
\right.
$$
$$
f(b_n) = \left\{
\begin{array}{ll}
(1 – \frac{1}{n})^2 & \quad 0 \leq n < 1 \\
2 & \quad 1 \leq n
\end{array}
\right.
$$
-Which implies that as $n\rightarrow 1,$ $f(a_n)\rightarrow 4$ and $f(b_n)\rightarrow 0$

-Since the sequences approach different points, the limit does not exist, thus, I can conclude that the function is discontinuous at $x=1$.

-Is this correct? Is this enough proof that $f(x)$ is discontinuous at $x=1$?

Answer 1

Calculate the one-sided limits at $x=1$:

1. $\displaystyle\lim_{x\to 1+}f(x)=\lim_{x\to 1+}2=2$
2. $\displaystyle\lim_{x\to 1-}f(x)=\lim_{x\to 1-}x^2=1$

Since they are not equal, $f$ is not continuous at $x=1$. The rest that you did has some mistakes (you do not calculate one-sided limits when you should) but more importantly is not necessary (and may be time consuming in an exam with limited time).