How can one show the following quotient has a limit of $1$?

Question

How can one show that the limit of the following is $1$?

$$\lim_{x\to 0}\frac{\frac{1}{1-x}-1}{x}=1$$

Answer 1

By definition of derivative with

$$f(x)=\frac{1}{1-x}\implies f'(x)=\frac{1}{(1-x)^2}$$

we have

$$\lim_{x\to 0}\frac{\frac{1}{1-x}-1}{x-0}=f'(0)=1$$

Answer 2

$$\frac{\dfrac{1}{1-x}-1}{x}=\frac{\dfrac{1-(1-x)}{1-x}}{x}=\frac{1}{1-x}\xrightarrow{x\to 0}1$$

Answer 3

Hint:

$$ \frac{\frac{1}{1-x}-1}{x}=\frac{\frac{1-(1-x)}{1-x}}{x}=\frac{x}{x(1-x)}=\frac{1}{1-x}. $$