Prove that the limit as $x \rightarrow 0$ of $\frac{x}{x-1} = 0$.

Question

Prove that the limit as $x \rightarrow 0$ of $\frac{x}{x-1} = 0$.

I'm thinking of putting it in this form $|x| \lt \epsilon|x-1|$ and bounding it somehow, but I'm really at a loss of how to continue. I tried letting $|x| \lt 1$ and then $-2 \lt x -1 \lt 0$ but I'm not sure how to continue as minimizing the right hand side would result in a $0$.

Anyone have any ideas?

Answer 1

Let $|x-1|>1/2$, e.g.

$-1/2 <x<1/2;$ or $|x|<1/2$.

Let $\epsilon$ be given.

Choose $\delta =\min (1/2, \epsilon/2);$

Then

$|x|<\delta$ implies

$|\frac{x}{x-1}| =|x|\frac{1}{|x-1|}<|x|\frac{1}{(1/2)}= $

$2|x| <2\delta \le 2(\epsilon/2)=\epsilon.$

Answer 2

Take a sufficiently small neighborhood of $0$, say $[-\frac12,\frac12]$ and you have $|x-1|\le\frac32$, and this is true for any smaller neighborhood. This ensures

$$|x|<\epsilon|x-1|\le\frac{3\epsilon}2.$$

Answer 3

Just apply the facts that the limit of a sum is the sum of limits if they exist, that the limit of a constant is the same constant and that the limit of a ratio $f(x)/g(x)$ is the ratio of the limits, given that they exist and the limit of $g(x)$ isn't $0$ (which is true for $g(x) = x-1$ as $x\to 0$).