Epsilon delta continuity of 1/x at x=1

[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.

Homework Statement

Show that 1/x is continuous at x=1

Homework Equations

If |x-a|<δ
Then |f(x)-f(a)|<ε

The Attempt at a Solution

x<δ+1
1/x<ε+1
(ε+1)x>1
1/(ε+1)<x<δ+1
*starts crying*

 

[HallsofIvy]

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&gt; 0 there exist \delta&gt; 0 such that if |x- a|&lt; \delta then |f(x)- L|&lt; \epsilon.

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

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

 

[pasmith]

You need to show that for all \epsilon &gt; 0 there exists \delta &gt; 0 such that if |x - 1| &lt; \delta then |1/x - 1| &lt; \epsilon.

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

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