Proving that a limit has been incorrectly evaluated, using the epsilon-delta definition of a limit

Question

Earlier, I was trying to prove that this evaluation of a limit is incorrect:

$$ \lim_{x\to2}x^2=4.01 $$

I tried using $\varepsilon=0.1$, but I could not find a corresponding value of $\delta$ that would prove that the above expression is incorrect. My reasoning for why this happened is as follows:

• Using $\varepsilon=0.1$ means that I need to values of $x$ where $x^2$ falls in the open interval $(3.91,4.11)$. Because this range includes the proper answer to the limit of $4$, I cannot use this value of $\varepsilon$. If I find a suitable value of $\delta$, then all the $x^2$ values will be sufficiently close to $4$, meaning that they will also be sufficiently close to $4.01$. E.g. $\delta=0.01$ means that the $x^2$ values will fall in the open interval $(3.9601,4.0401)$. Because this interval is 'encapsulated' by the interval $(3.91,4.11)$, there is no problem—even though the evaluation of the limit is incorrect.

Therefore, don't I need to pick values of $\varepsilon$ such that $x^2>4$ for all $x$, and proceed from there? I tried using $\varepsilon=0.005$, and that seemed to work.

Answer 1

but I could not find a corresponding value of $\delta$ that would prove that the above expression is incorrect

Well, that's because that isn't how it would work! If you want to negate the definition of a limit, you would have to show that there exists some $\epsilon > 0$ (this you can choose) for which no $\delta > 0$ can satisfy the required condition.

(This is the opposite of when you prove a limit. You start with an arbitrary $\epsilon>0$ and show that there exists some $\delta > 0$ such that ...)

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In this case, let us choose $\epsilon = 0.01 > 0$.
Now we show that given any $\delta > 0$, there exists $x \in (2 - \delta, 2 + \delta) \setminus \{2\}$ such that $|x^2 - 4.01| \ge \epsilon.$

(This would precisely show that $4.01$ is not the limit as this is the negation of the original definition.)

Well, simply choose $x = \max\{1, 2- \delta/2\}.$ Clearly, we have $0 < |x - 2| < \delta.$
If $x = 1$, then we clearly have $|x^2 - 4.01| = |3.01| \ge 0.1$.
If $x= 2 - \delta/2$, then we have that $0 < 2 - \delta/2 < 2$ and hence, $x^2 < 4$. Once again, the desired inquality follows.

Answer 2

Well, let's not kid ourselves. We know the limit is $2^2 =4$.

And if $\epsilon = 0.1$ then $|4.01 - 4| = 0.01 <0.1 = \epsilon$ we won't be able to find a contradiction because $\epsilon$ is too big.

(Let $\epsilon_\omega = 0.09$. Then we can find a $\delta$ so that $|x-\delta| \implies |4-x^2| < 0.09$. But $|4.01 - x^2| \le |4.01 - 4| + |4-x^2| < 0.01 + 0.09 = 0.1$ we can't have a contradiction.)

We need to find a contradiction with with a smaller $\epsilon$.

Try $0 < \epsilon < 0.009$

If $|4.01 - x^2| < 0.009$ then $4.01 < x^2 < 4.018$ and $\sqrt 4.01 < x < \sqrt{4.02}$ so $x > \sqrt{4.01} > 2$ and so for any $x$ where $2 < x < \sqrt{4.01}$ we will not have $|4.01 - x^2| < 0.009$. So no delta exist that assures our result.

Answer 3

Consider the problem based on your skills. Suppose you have learnt how to use the definition of limit in the positive manner (proving a number is limit) but have yet to learn to use it in negative manner.

Thus you have already established the following fact: for every $\epsilon >0$ there is a $\delta>0$ such that $$0<|x-2|<\delta\implies |x^2-4|<\epsilon $$

The current question asks us to prove that the $4$ in above statement can not be replaced by $4.01$ and this can be proved very very easily (since the hard part of proving the positive statement above has already been done).

Suppose on the contrary that a similar statement holds when $4$ is replaced by $4.01$. Taking $\delta$ as the minimum of the deltas obtained from these two statements (one proved and one assumed) we see that $$0<|x-2|<\delta\implies |x^2-4|<\epsilon, |x^2-4.01|<\epsilon $$ But then $$0.01=|4.01-4|\leq |4.01-x^2|+|x^2-4|<2\epsilon$$ We get an obvious contradiction if we choose $\epsilon=0.005$. This solves our problem without any messy dealing with decimals.

The above strategy essentially proves the uniqueness of limits.