[Math] What’s wrong with the epsilon delta proof here

calculusepsilon-deltafake-proofslimits

Claim: $$\lim_{x\to 1} \frac{100}{x} = 100$$

Proof: Let any $\epsilon > 0.\ |\frac{100}{x}-100| = 100 |\frac{1}{x}-1| = 100 \frac{|x-1|}{|x|}.$

If this term is smaller than epsilon in the interval where $0<|x-1|<\delta$, then we are done.

Assume $\delta=100.$ Then, $|x-1|<100, -99<x<101, -1/99<1/x<1/100$

Thus $100 \frac{|x-1|}{|x|} <\frac{ 100}{100*|x-1|}$

So if $0<|x-1|< \min\{\epsilon, 100\}=\delta,$ then $|\frac{100}{x}-100| < \epsilon$. QED

Note: Even if this is correct, my intent of asking this question was because I'm having doubts on the fundamental logic lying behind this proof… For some reason, I don't like the fact that we can let delta be something and then later "rectify" it after putting in a form of epsilon… Like I can let delta be less than 1/2 and get (I think) for delta = min{epsilon/200, 1/2} and that's different from what I got… I know that epsilon is any number, but it just seems weird. Did I misunderstand something?

I will be really grateful if someone can help me in any way understand this better.

Best Answer

You made a mistake in the implication that

$$ -99 < x < 101 \implies -\frac{1}{99} < \frac{1}{x} < \frac{1}{100} $$

When $A, B$ are positive numbers, you have (note the reversal of inequality signs)

$$ A < x < B \implies \frac{1}{A} > \frac{1}{x} > \frac{1}{B}$$

When $A,B$ are both negative numbers, you also have (the same thing)

$$ A < x < B \implies \frac{1}{A} > \frac{1}{x} > \frac{1}{B} $$

But when $A < 0 < B$, all you can say is

$$ A < x < B \implies \frac{1}{A} > \frac{1}{x} \mbox{ or } \frac{1}{B} < \frac{1}{x} $$

in particular, the absolute value

$$ \frac{1}{|x|} > \min( \frac{1}{|A|}, \frac{1}{|B|} )$$

is not bounded above a priori.

To address your query

For some reason, I don't like the fact that we can let delta be something and then later "rectify" it after putting in a form of epsilon

perhaps it would be clearer if the $\epsilon$-$\delta$ statements were written this way (I'll use the case of the limit as an example, but you can substitute similar statements into other definitions):

Given a function $f(x)$ and a point $x_0$, and a value $y$. We say that $\lim_{x\to x_0} f(x) = y$ if there "we can find" a function $\delta(\epsilon)$ (a function $\delta$ that depends on the variable $\epsilon$) such that the expression $|f(x) - y| < \epsilon$ holds true whenever $0 < |x-x_0| < \delta(\epsilon)$.

Rmk: I put "we can find" in quotes because in non-school mathematics (as practised by professional mathematicians), we usually cannot write down an explicit formula for the function $\delta(\epsilon)$. We just can demonstrate that such a function must exist.

In particular, doing an $\epsilon$-$\delta$ proof is like reducing a system of algebraic equations: you want to "solve" for the function $\delta(\epsilon)$ (subject to the above caveat).

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