Why the function has removable discontinuity

Question

$f(x) = \begin{cases} 2x-1 & \text{when }x<2 \\ 5 & \text{when }x=2 \\ \frac{1}{2}x + 2 & \text{when }x>2 \end{cases}$

I am learning Calculus But I can't seem to understand why this function has a removable discontinuity at $x=2$.

So far I have solved functions like this:

$f(x) = \frac{x^2-5x+6}{x-3} $

In such cases, I can clearly see that by factoring discontinuity at $x=3$ can be removed. But unfortunately, in the case of piecewise function above, I am not able to understand why discontinuity is removable at x = 2.

Answer 1

Removable discontinuity means that $\lim_{x\rightarrow 2^-}f(x)=\lim_{x\rightarrow 2+} f(x) $.

$\lim_{x\rightarrow 2^-}f(x)=2(2)-1=3$ and $\lim_{x\rightarrow 2^+}f(x)=2/2+1=3$.

Answer 2

Since both one side limits at $x=2$ are equal to $3$ we can remove the discountinuity at that point defining $f(2)=3$.

In that case we call that a removable discontinuity because the function becomes continuous at a point changing or defining its value at that point.

Another well known case is

$$f(x)=\frac{\sin x}{x}$$

not defined at $x=0$ which becomes continuous if we define $f(x)=1$ obtaining the $\operatorname{sinc}(x)$ function.