# How do you find discontinuity of a piecewise function?

In most cases, we should seem for a discontinuity at the subject-matter where a elementwise defined business changes its formula. You obtain entertain to seize one-sided limits individually past unanalogous formulas obtain direct depending on from which aspect you are approaching the subject-matter. Here is an issue.

Let us ponder where ##f## has a discontinuity.

##f(x)={(x^2 if x<1),(x if 1 le x < 2),(2x-1 if 2 le x):}##,

Notice that each element is a polynomial business, so they are true by themselves.

Let us see if ##f## has a discontinuity ##x=1##.

##lim_{x to 1^- }f(x)=lim_{x to 1^- }x^2=(1)^2=1##

##lim_{x to 1^+}f(x)=lim_{x to 1^+}x=1##

Since twain limits yield 1, ##lim_{x to 1}f(x)=1##

##f(1)=1##

Since ##lim_{x to 1}f(x)=f(1)##, there is no discontinuity at ##x=1##.

Let us see if ##f## has a discontinuity at ##x=2##.

##lim_{x to 2^- }f(x)=lim_{x to 2^- }x=2##

##lim_{x to 2^+}f(x)=lim_{x to 2^+}(2x-1)=2(2)-1=3##

Since the limits overhead are unanalogous, ##lim_{x to 2}f(x)## does not await.

Hence, there is a leap discontinuity at ##x=2##.

I trust that this was advantageous.