$\epsilon$-$\delta$ proof that $\lim\limits_{x \to 1} \frac{1}{x} = 1$.

Question

I'm starting Spivak's Calculus and finally decided to learn how to write epsilon-delta proofs.

I have been working on chapter 5, number 3(ii). The problem, in essence, asks to prove that

$$\lim\limits_{x \to 1} \frac{1}{x} = 1.$$

Here's how I started my proof,

$$\left| f(x)-l \right|=\left| \frac{1}{x} - 1 \right| =\left| \frac{1}{x} \right| \left| x - 1\right| < \epsilon \implies \left| x-1 \right| < \epsilon |x|$$

I haven't made any further progress past this point. Is it possible to salvage this proof? Should I try an alternate approach?

Answer 1

Update 2/19/2018: It appears that this answer has received a lot of attention, which I'm very glad to know about. When you're reading through this answer and you're trying to learn about $\delta$-$\epsilon$ proofs for the first time, I would recommend skipping the sections labeled Addendum. on your first read. Please let me know of any other clarifications that you would like with this answer.

---

Whenever I am doing a $\delta$-$\epsilon$ proof, I do some scratch work (note, this is NOT part of the proof) to figure out what to choose for $\delta$. I always tell students to think about the following:

1. What are you given?

2. What do you want to show?

In the definition of the limit, you are given an arbitrary $\epsilon > 0$ and you want to find $\delta$ such that $$0 < |x - 1| < \delta$$ implies $$\left|\dfrac{1}{x} - 1 \right| < \epsilon\text{.}$$ You have control over what to choose for your $\delta$ in this case. The idea of this $\delta$-$\epsilon$ proof is to work with the expression $|x - 1| < \delta$ and get $\left|\dfrac{1}{x} - 1 \right| < \epsilon$ at the end.

Let's do some scratch work (again, NOT part of the proof).

---

Scratch Work

Let's start with what we want to show for our scratch work (starting with what you want to show is bad to do $100\%$ of the time when you're doing proofs - again, this is scratch work and not actually part of the proof).

We want to show that $\left|\dfrac{1}{x} - 1 \right| < \epsilon$. Let's work backwards and try to turn the expression $\left|\dfrac{1}{x} - 1 \right|$ into some form of $|x-1|$.

So, note that $$\left|\dfrac{1}{x} - 1 \right| =\left|\dfrac{1-x}{x} \right| = \left|\dfrac{-(x-1)}{x} \right| = \left|\dfrac{x-1}{x} \right|$$ since $|y|=|-y|$ for all $y$ in $\mathbb{R}$.

The last expression can be rewritten as $\dfrac{\left|x-1 \right|}{\left| x \right|}$. Looking at this expression, we do have $|x-1|$ in the numerator, which is good. But we have that pesky $|x|$ in the denominator.

Since we do have control of what $|x-1|$ is less than (this is our $\delta$), let's choose a really convenient, small number to work with that is greater than $0$. Let's say $\delta = \dfrac{1}{2}$.

Well, if $|x - 1| < \dfrac{1}{2}$, then $$-\dfrac{1}{2} < x-1 < \dfrac{1}{2} \implies \dfrac{1}{2} < x < \dfrac{3}{2} \implies \dfrac{1}{2} < |x| < \dfrac{3}{2}\text{.}$$ So if we choose $\delta = \dfrac{1}{2}$, $\dfrac{1}{2} <|x| < \dfrac{3}{2}$.

Addendum. In many examples, $\delta$ is usually chosen to be $1$. Why did we elect not to do that in this case?

It's because it wouldn't work.

Intuitively, here's why it doesn't: when you consider the neighborhood of radius $1$ centered around $x = 1$, you get the interval $(0, 2)$. $f(x) = \dfrac{1}{x}$ doesn't have a finite limit at $x = 0$, so this makes $\delta = 1$ a bad choice.

This isn't the case if $\delta = 1/2$. The neighborhood of radius $1/2$ around $x = 1$ is $(1/2, 3/2)$. $f$ has limits at every $x$-value in the interval $(1/2, 3/2)$, including the endpoints.

In terms of the algebra, if we had chosen $\delta = 1$, the algebra wouldn't have worked out. We would've gotten $0 < x < 1$ and would not have been able to obtain a finite upper bound for $\dfrac{1}{x}$. That is, $$0 < x < 1 \implies 1 < \dfrac{1}{x} < \infty\text{.}$$ We do not have a finite upper bound for $\dfrac{1}{x}$ in this case, and hence why $\delta = 1$ will not work for this purpose.

If $\dfrac{1}{2} <|x| < \dfrac{3}{2}$, then $$\dfrac{2}{3} <\dfrac{1}{|x|} < 2$$ and $$\dfrac{1}{|x|} < 2 \implies \dfrac{\left|x-1 \right|}{\left| x \right|} < 2\left| x-1 \right|\text{.}$$ Now we have control over what $|x-1|$ is less than. So to get $\epsilon$, we choose $\delta = \dfrac{\epsilon}{2}$.

But, wait - didn't I say that we chose $\delta = \dfrac{1}{2}$ earlier? A simple solution would be to minimize $\delta$, i.e., make $\delta = \min\left(\dfrac{\epsilon}{2} , \dfrac{1}{2} \right)$.

Addendum. To see why $\delta = \min\left(\dfrac{\epsilon}{2} , \dfrac{1}{2} \right)$ works, suppose $\dfrac{\epsilon}{2} > \dfrac{1}{2}$, so that $\delta = \dfrac{1}{2}$. Then $\epsilon > 1$. Then $$\dfrac{\left|x-1 \right|}{\left| x \right|} < 2|x-1| < 2 \cdot \dfrac{1}{2} = 1 < \epsilon\text{.}$$

Now suppose $\dfrac{\epsilon}{2} \leq \dfrac{1}{2}$, so that $\delta = \dfrac{\epsilon}{2}$.

Then $$\dfrac{\left|x-1 \right|}{\left| x \right|} < 2|x-1| < 2 \cdot \dfrac{\epsilon}{2} = \epsilon\text{.}$$

In both cases, we have $\dfrac{\left|x-1 \right|}{\left| x \right|} < \epsilon$, as desired.

See also Why do we need min to choose $\delta$?.

So now we've found our $\delta$ and can use this to write out the proof.

---

The Proof

Proof. Let $\epsilon > 0$ be given. Choose $\delta := \min\left(\dfrac{\epsilon}{2} , \dfrac{1}{2} \right)$. Then $$\left|\dfrac{1}{x} - 1 \right| = \left|\dfrac{x-1}{x} \right| = \dfrac{\left|x-1 \right|}{\left| x \right|} < 2\left| x-1 \right|$$ (since if $|x - 1| < \dfrac{1}{2}$, $\dfrac{1}{|x|} < 2$) and $$2\left| x-1 \right| < 2\delta \leq 2\left(\dfrac{\epsilon}{2}\right) = \epsilon\text{. }\square$$

Success at last.

Addendum. Note that the end goal above was achieved, namely to show that $$\left|\dfrac{1}{x}-1\right| < \epsilon\text{.}$$

In the step $$2\left| x-1 \right| < 2\delta \leq 2\left(\dfrac{\epsilon}{2}\right) = \epsilon\text{,}$$ textbooks usually omit the step with the $\delta$ and just write $$2\left| x-1 \right| < 2\left(\dfrac{\epsilon}{2}\right) = \epsilon\text{.}$$

Addendum. It may seem that the note "(since if $|x - 1| < \dfrac{1}{2}$, $\dfrac{1}{|x|} < 2$)" may be an additional assumption added to the problem - i.e., that $\delta$ has to be $\dfrac{1}{2}$. This is not the case for the following reason: given $|x-1| < \delta$, we have $$|x-1| < \min\left(\dfrac{\epsilon}{2}, \dfrac{1}{2}\right)\text{.} $$ Obviously, if $\epsilon \geq 1$, we end up with $|x - 1| < \dfrac{1}{2}$, as stated above. But let's suppose that $\epsilon < 1$. Then $$|x - 1| < \dfrac{\epsilon}{2} < \dfrac{1}{2}$$ and you end up with $|x - 1| < \dfrac{1}{2}$, so the $\dfrac{1}{|x|} < 2$ implication holds in either case.

Answer 2

Take first $\delta =1/2$. Then $$||x|-1|<|x-1|<\frac 1 2$$ gives that $$-\frac 1 2 +1<|x|<\frac 1 2+1$$ $$\frac 1 2 <|x|$$so $$\frac 1 {|x|}<2$$

Now given $\epsilon$ take $$\delta= \min\left(\frac\epsilon 2,\frac 12\right)$$

---

In general, assume $\lim_{x\to a}f(x)=\ell \neq 0$ We have $$\left|\frac{1}{f(x)}-\frac 1{\ell}\right|=\frac{|f(x)-\ell|}{|f(x)||\ell|}$$

Take $\epsilon =|\ell|/2$ in the definition of $\lim_{x\to a}f(x)=\ell$ to get a $\delta_1>0$ such that $$-|\ell|/2<|f(x)|-|\ell|<|\ell|/2$$

to obtain the lower bound $$|f(x)|^{-1}<\frac 2{|\ell|}$$

Then

$$\left|\frac{1}{f(x)}-\frac 1{\ell}\right|<\frac{2|f(x)-\ell|}{ |\ell|^2}$$

whenever $0<|x-a|<\delta_1$. But given $\epsilon >0$ there exists $\delta_2>0$ such that

$$|f(x)-\ell|<\frac{|\ell|^2\epsilon}{2}$$

whenever $0<|x-a|<\delta_2$. Thus given $\epsilon >0$, we should take $$\delta=\min(\delta_1,\delta_2)$$ and we will have

$$\left|\frac{1}{f(x)}-\frac 1{\ell}\right|<\frac{2|\ell|^2\epsilon}{2 |\ell|^2}=\epsilon$$

Answer 3

I am pretty confident after one full year you must have finished this topic. But I have an alternative approach I would love to share here!!

$\displaystyle \lim_{x\to 1} \frac{1}{x} = 1$ tells us that there is some $\delta$ such that $|\frac{1}{x} – 1| < \epsilon$

$\implies \frac{|x – 1|}{|x|} < \epsilon$

Let’s assume $|x – 1| < \frac{1}{3}$ Therefore,

$2/3 < x < 4/3$

$\implies 2/3 < |x| < 4/3 \implies 3/2 < \frac{1}{|x|} < 3/4$

Let’s recall that $|x – 1| < \delta$ is also true

Since both $|x|$ and $|x – 1|$ are positive, we can consider that,

$|x – 1| < \delta$

$\frac{1}{|x|} < 3/4$

$\implies \frac{|x-1|}{|x|} < \frac{4\delta}{3}$

Considering the LHS, we know the LHS < \epsilon so we can let

$\frac{4\delta}{3} = \epsilon$

$\implies \delta = \frac{3\epsilon}{4}$

But this is under the assumption that $|x - 1| < 1/3$ therefore, the $\delta$ is actually,

$\delta = \min(1/3, \frac{3\epsilon}{4})$

Answer 4

Given $\epsilon>0$ we look for $\delta>0$ s.t. as if $|x-1|\leq \delta$ we have $\left| \frac{1}{x} - 1 \right|\leq \epsilon$.

If we add a further condition: $\delta\leq \frac{1}{2}$ and since $|x-1|\leq \delta$ we find by the triangle inequality $|x|\geq 1-\delta\geq \frac{1}{2}$ so we have $$\left| \frac{1}{x} - 1 \right| =\left| \frac{1}{x} \right| \left| x - 1\right| < \epsilon \iff \left| x-1 \right| < \epsilon |x|\Leftarrow |x-1|\leq \frac{\epsilon}{2}$$ so it suffices to choose $\delta=\min(\frac{\epsilon}{2},\frac{1}{2})$.

Answer 5

$\tag 1 |\frac{1}{x} - 1| < \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; -\varepsilon < \frac{1}{x} - 1 < \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; 1 -\varepsilon < \frac{1}{x} < 1 + \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; 1 -\varepsilon < \frac{1}{x} < 1 + \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; \frac{1}{1 + \varepsilon} < x < \frac{1}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; \frac{1}{1 + \varepsilon} < x < \frac{1}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; \frac{-\varepsilon}{1 + \varepsilon} < x - 1 < \frac{+\varepsilon}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

We are trying to find our $\delta \gt 0$ 'setup', $|x -1| \lt \delta$ contrained by

$\tag 2 \frac{-\varepsilon}{1 + \varepsilon} \le -\delta \lt x - 1 \lt +\delta \le \frac{+\varepsilon}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

so that $\text{(2)}$ implies $\text{(1)}$ (we can use our developed "$\text{iff }$-logic-chain" in reverse).

Setting

$\quad \delta = \frac{\varepsilon}{1 + \varepsilon}$

sets up the left side of $\text{(2)}$, but also takes care of the rights side since

$\quad \delta = \frac{\varepsilon}{1 + \varepsilon} \le \frac{\varepsilon}{1 - \varepsilon}$

Note: Here we don't use the 'min delta' approach. Using the same mechanical technique found here, we simply 'turn-the-crank'. As we work things out we realize that we want to constrain $\varepsilon$.

Instructional Example 1: You are challenged with an $\varepsilon$ that is greater than or equal to $1/2$.
Then set $\delta = 1/3$.

Note that if $\varepsilon = 1/500$, we take $\delta = 1/501$, a 'looser' number than the $\varepsilon/2 = 1/1000$ found in other (excellent) answers in this thread.