Question

Find the values of x (if any) at which f(x)= x+1/x^2-1 is not continuous. If so, is the discontinuity removable?

Answer 1

$f(x)=\dfrac{x+1}{x^2-1}$ has two points of discontinuity at $x=\pm1$, where $x^2-1=0$.

We have

$\dfrac{x+1}{x^2-1}=\dfrac{x+1}{(x+1)(x-1)}$

so as long as $x\neq-1$, we can cancel $x+1$ and be left with $\dfrac1{x-1}$. Then as $x\to-1$, we have

$\displaystyle\lim_{x\to-1}f(x)=\lim_{x\to-1}\frac1{x-1}=-\frac12$

The limit exists, so the discontinuity at -1 is removable (so definitely not C or D).

Meanwhile,

$\displaystyle\lim_{x\to1}f(x)=\lim_{x\to1}\frac1{x-1}$

does not exist, so the discontinuity at 1 is non-removable, making A the correct answer.