# Epsilon delta continuity of 1/x at x=1

1. Feb 28, 2013

### Avatrin

This isn't really homework; It's just something that has been bothering me ever since I first learned calculus because I suck at epsilon-delta proofs.
1. The problem statement, all variables and given/known data
Show that 1/x is continuous at x=1

2. Relevant equations
If |x-a|<δ
Then |f(x)-f(a)|<ε

3. The attempt at a solution
x<δ+1
1/x<ε+1
(ε+1)x>1
1/(ε+1)<x<δ+1
*starts crying*

2. Feb 28, 2013

### HallsofIvy

Staff Emeritus
You appear to have a confused idea of what limits are. You cannot just choose δ and
ε independently. The definition of "limit" is
$\lim_{x\to a} f(x)= L$ if and only if given $\epsilon> 0$ there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- L|< \epsilon$.

That is, $\delta$ will typically depend upon $\epsilon$. The simplest way to prove a limit is to start from $|f(x)- L|<\epsilon$ and use that to derive a value for $\delta$. As long as every step is "reversible that shows we can from $|x- a|< \delta$ to $|f(x)- L|< \epsilon$.

Here, f(x)= 1/x, a= 1, and L= 1. So, given $\epsilon> 0$ we want to get $|1/x- 1|< \epsilon$ (notice the absolute value- you didn't have that). That is the same as $-\epsilon< (1/x)- 1< \epsilon$ which is the same as $-\epsilon< (1- x)/x< \epsilon$. As long as x is close to 1 ($\delta$< 1) x> 0 so $-x\epsilon< 1- x< x\epsilon$.

3. Feb 28, 2013

### pasmith

You need to show that for all $\epsilon > 0$ there exists $\delta > 0$ such that if $|x - 1| < \delta$ then $|1/x - 1| < \epsilon$.

We will clearly need $0 < \delta < 1$ (or else 0 will be in our interval). Now 1/x is strictly decreasing for $x > 0$ (this is a consequence of $\mathbb{R}$ being an ordered field), so for $0 < 1 - \delta < x < 1 + \delta$ the following inequality is true:
$$\frac{1}{1 + \delta} < \frac1x < \frac{1}{1 - \delta}$$

So if given $\epsilon > 0$ we can find $\delta$ such that
$$1 - \epsilon < \frac{1}{1 + \delta} < \frac1x < \frac{1}{1 - \delta} < 1 + \epsilon$$
then we will be done. Note that if $\epsilon \geq 1$ then the left-most inequality is satisfied for any $\delta > 0$.