What are the removable and non-removable discontinuities, if any, of #f(x)=2/(x+1)#?

Question

Aurora Alvarado · Accepted Answer

There is a non-removable discontinuity when #x=-1#.

The only discontinuity you can have with a quotient of continuous functions is when the denominator is #0#. In your case, when #x+1 = 0 \iff x = -1#. If you want to know if the discontinuity is removable, you can just compute the limit and see if it exists : #\lim_( x \rightarrow -1) f(x) = \lim_( x \rightarrow -1) 2/(x+1) = +- oo#. Therefore, it is not removable.

Cameron Alexander · Answer

undefined The function f(x) = 2/(x+1) has a removable discontinuity at x = -1. There are no non-removable discontinuities.

Charles Anctil · Answer

undefined The function $ f(x) = \frac{2}{x + 1} $ has a removable discontinuity at $ x = -1 $ because it results in a zero denominator, making the function undefined at that point. However, the function can be redefined at $ x = -1 $ by removing the singularity and defining $ f(-1) $ as a limit, resulting in a continuous function.

There are no non-removable discontinuities in the given function since it is defined for all real numbers except $ x = -1 $, where the function has a removable discontinuity. Therefore, there are no other points where the function is undefined or exhibits a non-removable discontinuity.