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

1 Answer
Apr 2, 2016

Answer:

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

Explanation:

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.