finding points of discontinuity of the function $f(x) = \frac{\frac{1}{x} - \frac{1}{x+1}}{ \frac{1}{x-1} - \frac{1}{x}}$.

Question

finding points of discontinuity of the function $f(x) = \frac{\frac{1}{x} - \frac{1}{x+1}}{ \frac{1}{x-1} - \frac{1}{x}}$.

My answer is:

After simplifying the function it becomed $$f(x) = \frac{x-1}{x+1}$$ and I found that the function is undefined at $x = -1$ and limit $f(x)$ as x tends to (-1) is $-\infty$, hence the function has essential discontinuity at this point.

Am I correct?

Answer 1

Algebra has allowed you to change the formula of the function, simplifying it.

But there was a cost: not only did you change the formula, you changed the function itself. Before applying the algebra and changing the formula, it is clear from the formula that plugging in $x=0$ and $x=1$ and $x=-1$ lead to zeroes in denominators which make no sense. So those three values of $x$ are not in the domain. After changing the formula, you can still see that $x=-1$ is not in the domain, but those algebraic alterations have hidden an important reality: you still cannot plug $x=0$ and $x=1$ into the original formula for the function.

Answer 2

we have to take two term into account,they are $\frac{1}{x},\frac{1}{x+1}$ to be defined at those points it is necessary to $x\ne0$ and $x\ne-1$ you can also think,why am I not taking $\frac{1}{x-1}$ part into account.Let's see, if $x\to1$,then the function becomes like $$\lim_{k\to \infty}\dfrac{something}{k}=0$$ so,it doesn't affect the function.so,for the original function the points of discontinuity are $x=-1,0$

NOTE: if you simplify the function then you are removing the removable discontinuous points.the function has a removable discontinuity at point $x=0$.