# Why the function has removable discontinuity

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

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.

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

• so unlike the rational function in my question, in the case of piecewise function, there is no factoring involved?
– Cody
Oct 28, 2018 at 5:35
• one more thing how I can redefine the continuity at $x=2$?
– Cody
Oct 28, 2018 at 5:44
• By setting $f(2)=3$. Oct 28, 2018 at 5:48

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.