Question

Please explain how they plugged the removable discontinuity in this question.

Answer 1

In the limit

`\displaystyle \lim_(x\to c) f(x)`

we're interested in the value that
`f(x)` converges to as
`x` gets closer to
`c`. So in fact
`x\\eq c`.

In the given example,
`f(x)` is factorized to reveal a common factor of
`x-1` in the numerator and denominator. We have
`x\\eq1` if
`x\to1`, so
`x-1\\eq0` so we can simplify

`(x-1)/(x-1) = 1`

and *remove* the discontinuity.

Then

`\displaystyle \lim_(x\to1) (2(x-1))/((x+1)(x-1)) = \lim_(x\to1) \frac2{x+1} = \frac2{1+1} = 1`