What are the removable and non-removable discontinuities, if any, of #f(x)=x^2/absx#?

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Answer 1 · 2015-11-02T12:33:12

#f# has a removable discontinuity at #0#.

Recall that #absx = {(x, "if",x >= 0),(-x, "if",x < 0) :}#

So #f(x) = {(x^2/x, "if",x > 0),(x^2/-x, "if",x < 0) :}#

Notice that #f# is not defined for #x=0#, so #f# is not continuous at #0#.

We can simplify the definition of #f#, to get

#f(x) = {(x, "if",x > 0),(-x, "if",x < 0) :}#

This is simply the absolute value function except that it is left undefined at #0#.

We know (I think) that the absolute value function is continuous, so #f# is continuous everywhere except #0#.

Although #f(0)# is not defined, we can see that #lim_(xrarr0)f(x)# exists. (It equals #0#), so the discontinuity at #0# is removable.