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 $x\neq c$.

In the given example, $f(x)$ is factorized to reveal a common factor of $x-1$ in the numerator and denominator. We have $x\neq1$ if $x\to1$, so $x-1\neq0$ so we can simplify

$\dfrac{x-1}{x-1} = 1$

and *remove* the discontinuity.

Then

$\displaystyle \lim_{x\to1} \frac{2(x-1)}{(x+1)(x-1)} = \lim_{x\to1} \frac2{x+1} = \frac2{1+1} = 1$