Let's consider the function
f(x)={3−x,x≥2x+1,x<2 Obviously, it is continuous at any point except x=2. It is necessary to analyze the function at the point x=2.
It's obvious that
x→2+0limf(x)=x→2+0lim3−x=1 and
x→2−0limf(x)=x→2−0limx+1=3 So, both one-sided limits exist and are finite and not equal. It means that function f(x) has discontinuity of the first kind at x=2.
The graph
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