# ϵ-δ proof that $\lim_{x\to 2} \frac{1}{x} = \frac{1}{2}$

I've been doing some practice and for some reason I've been baffled by this since I get stuck on some point every time I restart. I've been doing some practice and for some reason I've been baffled by this since I get stuck on some point every time I restart.

## What I did

Prove $\lim_{x\to 2} \frac{1}{x} = \frac{1}{2}$ using $\epsilon - \delta$ defitinion:

$\left\lvert\frac{1}{x} - \frac{1}{2}\right\rvert < \epsilon$

$\left\lvert\frac{2-x}{2x}\right\rvert < \epsilon$

$\left\lvert\frac{x-2}{2x}\right\rvert < \epsilon$

If $\left\lvert x-2\right\rvert < \delta = \frac{1}{2}$

$\frac{-1}{2}<x-2<\frac{1}{2}$

$\frac{3}{2}<x<\frac{5}{2}$

Therefore $\frac{2}{3} > \frac{1}{x} > \frac{2}{5}$

$\frac{1}{\left\lvert x \right\rvert} < \frac{2}{3}$

$\frac{\left\lvert x-2 \right\rvert}{2\left\lvert x \right\rvert} < \frac{2\left\lvert x-2 \right\rvert}{3*2} = \frac{x-2}{3} = \delta$

$x-2 = 3\delta = \epsilon$

$\delta = \frac{\epsilon}{3}$

And after this point I can't find anything.

• The limit is actually 1/2. – Augustin Oct 9 '16 at 14:36
• Why don't you show us one or more of your failed attempts? You're expected to show your effort or thoughts before asking for help onMath.SE – GFauxPas Oct 9 '16 at 14:37
• Use the fact that you can assume $x>1$, so $1/x\le 1$. – egreg Oct 9 '16 at 14:44
• I am going to add what I did in a bit. – Eren O. Oct 9 '16 at 14:48
• With the added material this is a much better question. It shows you already have most of the pieces of the proof, so an answer doesn't need to explain all those pieces to you; it's mainly a matter of how to put it together in a valid proof. – David K Oct 9 '16 at 16:27

What you tried is actually a mostly reasonable sequence of steps to discover the proof you need, except for a couple of significant problems.

The first error (though it's only going to cause a problem if you try to use it again later) is setting $\delta = \frac12$. Clearly you can't just set $\delta$ equal to $\frac12$ in all cases; that will allow choices of $x$ such as $x = 2 - \frac14$, which says $\frac1x = \frac47$, and that just won't do when $\epsilon = 0.001$.

What you should have written (and possibly what you really were thinking of) is that you want $\delta \leq \frac12$.

There's also some confusion later, where you wrote "$\frac{2\lvert x-2 \rvert}{2\lvert x \rvert} < \frac{2\lvert x-2 \rvert}{3\cdot 2} = \frac{x-2}{3} = \delta$." The first inequality was correct, and the cancellation of the $2$ was fine, but in general $x - 2 \neq \lvert x-2 \rvert$; and keep in mind that the goal is to show that $\left\lvert\frac{1}{x} - \frac{1}{2}\right\rvert < \epsilon$, not to show $\left\lvert\frac{1}{x} - \frac{1}{2}\right\rvert < \delta$. So that line should have retained the absolute value signs around $x - 2$ and should have set the result less than $\epsilon$ rather than equal to $\delta$.

What you would discover then is that you need $\lvert x-2 \rvert < 3\epsilon$, which is satisfied when $\lvert x-2 \rvert < \delta$ and $\delta \leq 3\epsilon$.

In other words, you don't need $\delta$ to be as small as $\frac\epsilon3$. On the other hand, one thing to keep in mind about $\epsilon$-$\delta$ proofs is that you can never go wrong if you set $\delta$ smaller than it needs to be. The only real mistake you can make in choosing $\delta$ is to allow it to be too large.

So let's go with your original choice of $\frac\epsilon3$ and see what happens. But again, we don't set $\delta$ equal to $\frac\epsilon3$. If we did that, then when $\epsilon = 3$ (for example) we would have to set $\delta = 1$, but then $\delta \not< \frac12$, which spoils all the conclusions you reached by assuming $\delta < \frac12$. So instead of an equation, set $\delta \leq \frac\epsilon3$.

Now if you assert that $\delta \leq \frac12$ and you also assert that $\delta \leq \frac\epsilon3$, you can combine these into one assertion: $$\delta \leq \min\left\{\frac12, \frac\epsilon3 \right\}$$ where $\min\{ \ldots \}$ says to take the smallest value of the things inside $\{ \ldots \}$. You could also use $<$ instead of $\leq$ if you prefer.

Now to write the actual proof, the idea is to take all the bits and pieces of calculation that you made and put them in a logical sequence that starts from legitimate assumptions and ends with the desired conclusion. The legitimate assumptions in this case are that some value of $\epsilon$ has been given such that $\epsilon > 0$, and that given this value of $\epsilon$, a value of $\delta$ is chosen such that $\delta \leq \min\left\{\frac12, \frac\epsilon3 \right\}$.

By choosing $\delta$ this way, $\delta \leq \frac12$ and $\delta \leq \frac\epsilon3$. Then rather than, "If $\lvert x - 2 \rvert < \delta = \frac12$," you could write (as the next steps of your proof)

If $\lvert x - 2 \rvert < \delta$ then $\lvert x - 2 \rvert < \delta \leq \frac12$, so $\lvert x - 2 \rvert < \frac12$.

Your four lines after that, ending in $\frac{1}{\left\lvert x \right\rvert} < \frac23$, then are fine, and I think you could include them in the proof exactly as written. The line you wrote after that has some material you can use next in the proof, but it needs some correction; also, this might be a good place to bring in the equation $\left\lvert\frac1x - \frac12\right\rvert = \left\lvert\frac{2-x}{2x}\right\rvert$, which the first few lines of your attempt indicate you found. So you then have

$\left\lvert\frac1x - \frac12\right\rvert = \left\lvert\frac{2-x}{2x}\right\rvert = \frac{2\lvert x-2 \rvert}{2\lvert x \rvert} < \frac{2\lvert x-2 \rvert}{3\cdot 2} = \frac{\lvert x-2 \rvert}{3}$.

You could almost write $\frac{\lvert x-2 \rvert}{3} < \epsilon$ at the end of this, because it's true, but I'd recommend you explain how we know it's true rather than just assert it. That's when you can remind us that we assumed $\lvert x-2 \rvert < \delta$ and $\delta \leq \frac\epsilon3$, and use those facts to show that $\frac{\lvert x-2 \rvert}{3} < \epsilon$ and therefore $\left\lvert\frac1x - \frac12\right\rvert < \epsilon$.

If you finish the proof that way, you'll should be able to see that $\delta \leq \min\left\{\frac12, \frac\epsilon3 \right\}$ sets $\delta$ smaller than needed. In fact, you could have used $\delta \leq \min\left\{\frac12, 3\epsilon \right\}$, and that would have been good enough to finish the proof. But you could also omit the factor of $3$ entirely, and just write $\delta \leq \min\left\{ \frac12, \epsilon \right\}$, exploiting the idea that it's OK to choose $\delta$ smaller than necessary.