We say that $f(x)$ has limit $L$ (or $f(x)$ converges to $L$) when $x$ tends to $a$, denoted by $\lim_{x \to a} f(x) = L$, if
- formally, $\forall \varepsilon > 0, \exists \delta > 0, \forall x, \quad 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon$;
- less formally, for any given tolerance $\varepsilon > 0$, there is a sufficiently small threshold $\delta > 0$ such that whenever $x$ is closer to $a$ than this threshold, then $f(x)$ is closer to $L$ than the given tolerance;
- informally, $f(x)$ can be arbitrarily close to $L$, provided that $x$ is sufficiently close to $a$.
It's like playing a 2-player game: the first player give an $\varepsilon > 0$, and the second has to respond using a $\delta > 0$.
Proving $\lim_{x \to a} f(x) = L$ is to show that a response $\delta$ exists for every $\varepsilon$.
Example of non-existing limit: The function $f(x) = \dfrac{1}{x}$ has no limit as $x \to 0$.
We may show this by contradiction: Suppose that $\lim_{x \to 0} \dfrac{1}{x} = L$ for some $L \in \mathbb{R}$.
By definition, $$\forall \varepsilon > 0, \exists \delta > 0, \forall x, \quad 0 < |x| < \delta \implies \left|\frac{1}{x} - L\right| < \varepsilon.$$
This is a claim which holds for all $\varepsilon > 0$, in particular it holds for (let us say) $\varepsilon = 1$ (specialization of the quantifier $\forall$). Hence $$\exists \delta > 0, \forall x, \quad 0 < |x| < \delta \implies \left|\frac{1}{x} - L\right| < 1.$$
(In words, there exists a $\delta > 0$ such that $\left|\dfrac{1}{x} - L\right| < 1$ whenever $0 < |x| < \delta$). Let us show that the existence of such a $\delta$ is contradictory. Indeed, if such $\delta$ existed, let us take $x > 0$ which is sufficiently small such that $x < \frac{1}{|L| + 1}$ and $x < \delta$, says $$x = \frac{1}{2}\min\left\{\delta, \frac{1}{|L| + 1}\right\}.$$ On the one hand, since $0 < x < \delta$, the definition of $\delta$ gives $$\left|\dfrac{1}{x} - L\right| < 1.$$
On the other hand, $$\left|\dfrac{1}{x} - L\right| \ge \frac{1}{|x|} - |L| > |L| + 1 - |L| = 1.$$ This contradiction conludes the proof.
Bài viết đã được chỉnh sửa nội dung bởi nmlinh16: 29-01-2023 - 16:12
