How do you find the x values at which #f(x)=1/(x^2+1)# is not continuous, which of the discontinuities are removable?

Answer 1

Luke Alvarez

#f(x)# is continuous everywhere

I am assuming that #x# is real (ie #x in RR#) and we are not talking about a complex variable function (#x in CC#).

Th denominator is positive for all real numbers (we write this as #x^2+1>0 AA x in RR#

As such #f(x)# is continuous everywhere

graph{1/(x^2+1) [-30, 30, -1.5, 1.5]}

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Answer 2

Charles Anctil

The function f(x) = 1/(x^2+1) is not continuous at x = ±i, where i is the imaginary unit. Both of these discontinuities are non-removable.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.