Simplifying expression and its condition

Question

Simplify the expression and decide for which 'x' it makes sense

$$\frac{1-\frac{1}{1-\frac{1}{x-1}}}{1+\frac{1}{1+\frac{1}{x+1}}}$$

what is simplification and conditions of x?

Answer 1

We have

$$\frac{1-\frac{1}{1-\frac{1}{x-1}}}{1+\frac{1}{1+\frac{1}{x+1}}} =\frac{1-\frac{1}{\frac{x-2}{x-1}}}{1+\frac{1}{\frac{x+2}{x+1}}} =\frac{1-\frac{x-1}{x-2}}{1+\frac{x+1}{x+2}} =\frac{\frac{-1}{x-2}}{\frac{2x+3}{x+2}}=-\frac{x+2}{(2x+3)(x-2)}$$

with $x\neq 1,-1,2,-2,-\frac32$

Answer 2

We have $$1+\frac{1}{x+1}=\frac{x+1+1}{x+1}=\frac{x+2}{x+1}$$ $$1+\frac{1}{\frac{x+2}{x+1}}=\frac{x+1+x+2}{x+2}=\frac{2x+3}{x+2}$$ $$1-\frac{1}{x-1}=\frac{x-2}{x-1}$$ $$1-\frac{1}{\frac{x-2}{x-1}}=-\frac{1}{x-2}$$ Can you proceed?

Answer 3

We have: $$1-\frac{1}{1-\frac{1}{x-1}} = 1-\frac{x-1}{x-2} = \frac{x-2-x+1}{x-2}=\frac{-1}{x-2}$$

and

$$1+\frac{1}{1+\frac{1}{x+1}} = 1+\frac{x+1}{x+2} = \frac{2x+3}{x+2}$$

so

$$\frac{1-\frac{1}{1-\frac{1}{x-1}}}{1+\frac{1}{1+\frac{1}{x+1}}} = \frac{\frac{-1}{x-2}}{\frac{2x+3}{x+2}} = -\frac{x+2}{(x-2)(2x+3)}$$

and it makes sense $\forall x \not= 2, -2, 1, -1$ and $\dfrac{-3}{2}$ which is where the denominator is zero.

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$\underline{\text{Side note}}$: However, $2$ and $\dfrac{-3}{2}$ are the only real poles, while the others are removable singularities, so we can extend by continuity.