Understanding the precise definition of limit

Question

I'm attempting to understand the exact definition of a limit, but I'm struggling to do so.

For instance, why is it incorrect to say that the limit of x as it approaches 2 is equal to 10? $\lim_{x\rightarrow 2}x=10$

$$\forall\epsilon>0,\exists\delta>0,0<|x-c|<\delta\longrightarrow|f(x)-L|<\epsilon$$ I choose $\delta=100$. So why the following statement is not correct?

$$0<|x-2|<100\longrightarrow|x-10|<\epsilon,\quad\forall \epsilon>0$$

Answer 1

You are implicitly asking two questions:

Why is not the case that $\lim_{x\to 2}x=10$?

Why is not true that for every $\varepsilon>0$, $$ 0<|x-2|<100\implies |x-10|<\varepsilon \, ? $$

Let me address the second question first. Notice that in the statement $$ \forall\epsilon>0:0<|x-2|<100\implies |x-10|<\varepsilon $$ the symbol $x$ is not quantified; that is, we aren't explicitly told whether to read the sub-formula $0<|x-2|<100\implies |x-10|<\varepsilon$ as being true for all values of $x$ or just true for some value of $x$. However, in such a situation, it is common mathematical convention to assume the first interpretation. So the statement to which you refer is actually an abbreviation of $$ \forall \varepsilon>0:\forall x:0<|x-2|<100\implies |x-10|<\varepsilon \, . $$ Now you ask why this is false: well, it's because we can find values of $\varepsilon>0$ and $x$ such that $0<|x-2|<100$ does not imply $|x-10|<\varepsilon$. For instance, if $\varepsilon=1$ and $x=50$ then $0<|x-2|<100$, but it is not the case that $|x-10|<\varepsilon$. Therefore, the implication $0<|x-2|<100\implies |x-10|<\varepsilon$ does not hold for all values of $\varepsilon>0$ and $x$.

Now let's look at your first question. In general, to say that $\lim_{x\to a}f(x)\neq l$ means $$ \exists \varepsilon>0:\forall \delta>0:\exists x:(0<|x-a|<\delta)\land \bigl(|f(x)-l|\ge \varepsilon\bigr) \, . $$ Here, $a=2$, $f(x)=x$, and $l=10$, so we are trying to prove that $$ \exists\varepsilon>0:\forall \delta>0:\exists x:(0<|x-2|<\delta)\land \bigl(|x-10|\ge \varepsilon\bigr) \, . $$ This is a good exercise. Hint: $\varepsilon=1$ suffices to prove the statement. So try showing that $$ \forall \delta >0:\exists x:(0<|x-2|<\delta)\land \bigl(|x-10|\ge 1\bigr) $$ Earlier, we found that if $\delta=100$, then there is a corresponding $x$, namely $50$, such that $0<|x-2|<\delta$ and $|x-10|\ge 1$. But you need to prove that for an arbitrary $\delta>0$, there is an $x$ (dependent on $\delta$) such that $0<|x-2|<\delta$ and $|x-10|\ge 1$.

An alternative method of showing that $\lim_{x\to 2}x\neq 10$ is to use the uniqueness of limits: this is the theorem which states that if $f(x)\to l$ as $x\to a$, but also $f(x)\to l'$ as $x \to a$, then $l=l'$. Indeed, it is because of this theorem that the notation $\lim_{x \to a}f(x)$ makes sense to begin with. Since $x\to2$ as $x\to2$ (prove this!), it cannot be the case that $x\to10$ as $x\to2$.

Answer 2

Other answers and comments here are trying to clear up your misunderstanding. I am responding to this particular question:

So why the following statement is not correct?

$$0<|x-2|<100\longrightarrow|x-10|<\epsilon,\quad\forall \epsilon>0$$

Since that statement says "$\forall \epsilon>0$" it should be true for any particular $\epsilon$. So let's consider $\epsilon = 5$. The left side of the implication (the hypothesis) is $0<|x-2|<100$. That's surely true for $x=101$. But the right side is then |101-10|<5, which is false.