Proving $ \lim_{x\to 1}1-\frac{2}{x}=-1 $ using epsilon delta

Question

How can I prove this limit using epsilon delta without making restrictions to a delta?

$$ \lim_{x\to 1}1-\frac{2}{x}=-1 $$

Answer 1

Given $\epsilon>0$, we look for $\delta $ such that

$$0 <|x-1|<\delta\implies |1-\frac {2}{x}+1|<\epsilon $$

or $$0 <|x-1|<\delta\implies 2|\frac {x-1}{x}|<\epsilon $$

As $x $ goes to $1$, we can suppose that $x $ is not far from $1$, for example we can assume that $$|x-1|<\color {red}{\frac {1}{2}} $$ or $$\frac {1}{2}<x <\frac {3}{2}$$

and $$\frac {2}{3}<\frac {1}{x}<2$$

With this additional condition, we will look for $\delta>0$ such that $$|x-1|<\delta\implies 2|\frac{x-1}{x}|<4|x-1|<\epsilon $$

So you take $$\delta=\min (\color {red}{\frac {1}{2}},\frac {\epsilon}{4}) $$

Answer 2

Hint:

Solve the inequality $$ \left|1-\frac{2}{x}+1 \right|<\epsilon $$

and verify that the solution is a neighborough of $1$.

Answer 3

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

You have to take an appropriate $\delta$ to keep $x$ ''away'' from $0$

So you must restrict $\delta$ or in case you solve the inequality you have to make some restrictions to $\epsilon$ a priori.

Take $\delta=\min\{ \frac{\epsilon}, \frac{1}{2}\}$ and you are fine.

Answer 4

Given $\epsilon>0$ you want to find $\delta>0$ such that $|(1-\frac2x)-(-1)|<\epsilon$ when $|x-1|<\delta$.

Now, $|(1-\frac2x)-(-1)| = |2 - \frac2x| = 2|1-\frac1x| = \frac{2}{|x|} |x-1| < \frac{2}{|x|} \delta$ when $|x-1|<\delta.$ But we need to limit $\frac{2}{|x|}.$ To do this we make sure that $|x-1|<\frac12$ because then $x>\frac12$ so $\frac{2}{|x|} < \frac{2}{1/2} = 4.$

Thus, if we take $\delta < \min(\frac12, \frac14\epsilon)$ we get $|(1-\frac2x)-(-1)| < 4 \cdot \frac14\epsilon = \epsilon.$