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11.25 Continuous

Interpretation

If \(f : X \to \mathbb{R}\) and \(x_0\) is a limit point of \(X\), then \(\lim_{x \to x_0} f(x) = l\) if:

\(\forall \epsilon > 0, \exists \delta > 0\) such that \(0 < |x - x_0| < \delta \implies |f(x) - l| < \epsilon\)

Equivalent to have \(\delta = W(\varepsilon)\)

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Cosine function

\(\forall x > 0\) image​

\(\delta \text{ such that }\)If you walk from \((0, 0)\) a distance \(\delta \, \text{cm}\), you will have \(P\) above your head.

Theorem

Let \(X \subseteq \mathbb{R}\), \(f: X \to \mathbb{R}\), and let \(x_0\) be a limit point of \(X\). The following statements are equivalent:

  1. \(\lim_{x \to x_0} f(x) = l\)

  2. For all sequences \((x_n)\), \(x_n \in X\) such that \(x_n \neq x_0\) and \(\lim_{n\to\infty}x_{n}=x_0\), we have \(\lim_{x\to x_0}f(x)=\lim_{n\to\infty}f(x_{n})=l\).

Proof

\(1 \implies 2\)

We know \(\forall\varepsilon_1>0\), \(\exists\delta\): \(0<|x-x_0|<\delta\implies|f(x)-l|<\varepsilon_1\) (*)

And \(\forall\varepsilon_2>0,\exists n_0:n>n_0,\left|x_{n}-x_0\right|<\varepsilon_2\)

We need \(\forall\varepsilon>0,\exists n_0:n>n_0,\left|f\left(x_{n}\right)-l\right|<\varepsilon\)

Let \(\varepsilon_2=\delta\), we have \(\left|x_{n}-x_0\right|<\delta\), and let \(\varepsilon_1=\varepsilon\) we have (use (*)) \(\forall\varepsilon>0\), \(\exists\delta\): \(0<|x_{n}-x_0|<\delta\implies|f(x_{n})-l|<\varepsilon\)

\(2 \implies 1\) By contradiction.

Assume (1) is false: \(\exists \epsilon_0 > 0\) such that \(\forall \delta > 0\), \(\exists x_\delta\) such that \(0 < |x_\delta - x_0| < \delta\) but \(|f(x_{\delta})-l|\geq\epsilon_0\).

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Let \(\delta = \frac{1}{n}\). \(\exists x_{1/n}\) such that: \(|x_{1/n} - x_0| < \delta\) but \(|f(x_{1/n}) - l| > \epsilon_0\)

Define \(a_n = x_{1/n}\) for all \(n\).

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We know:

  1. \(|a_n - x_0| < \frac{1}{n} \implies x_0 - \frac{1}{n} < a_n < x_0 + \frac{1}{n}\)​\(\implies a_n \to x_0\) as \(n \to \infty\) by sandwich

  2. But \(|f(a_n) - l| > \epsilon_0\) \(\forall n\)​\(\implies \lim_{n \to \infty} f(a_n) \neq l\)
    Contradiction!

Analysis: We can imply the limit of sequence from the limit of function, thus we should always start from function, thus we use contradiction here and put a sequence

Corollary

Let \(f: X \to \mathbb{R}\), \(x_0\) be a limit point of \(X\).

If there exist \((x_n)\) and \((y_n)\) in \(X\) such that \(x_n, y_n \neq x_0\) and \(\lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n = x_0\), but \(\lim_{n \to \infty} f(x_n) \neq \lim_{n \to \infty} f(y_n)\)

Then \(\lim_{x \to x_0} f(x)\) does not exist.

Proof

If \(\lim_{x\to x_0}f(x)=l\implies\lim_{n\to\infty}f(x_{n})=l=\lim_{n\to\infty}\left(y_{n}\right)\),

Contradiction

Example

Show that \(\lim_{x \to 0} \sin\left(\frac{1}{x}\right)\) does not exist

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\(x_n = \frac{1}{2n\pi} \in X\), we have \(\lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{1}{2n\pi} = 0\).

Then \(f(x_n) = f\left(\frac{1}{2n\pi}\right) = \sin\left(\frac{1}{\frac{1}{2n\pi}}\right) = \sin(2n\pi) = 0.\)

\(y_n = \frac{1}{2n\pi + \frac{\pi}{2}} \in X\), we have \(\lim_{n \to \infty} y_n = \lim_{n \to \infty} \frac{1}{2n\pi + \frac{\pi}{2}} = 0.\)

Then \(f(y_n) = f\left(\frac{1}{2n\pi + \frac{\pi}{2}}\right) = \sin\left(\frac{1}{\frac{1}{2n\pi + \frac{\pi}{2}}}\right) = \sin\left(2n\pi + \frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1.\)

Since \(\lim_{n \to \infty} f(x_n) \neq \lim_{n \to \infty} f(y_n)\), the limit \(\lim_{x \to 0} f(x)\) does not exist.

Proposition

If \(P(x) = a_n x^n + \dots + a_0\) and \(Q(x) = b_m x^m + \dots + b_0\) are polynomials, and \(Q(x_0) \neq 0\), then \(\lim_{x \to x_0} \frac{P(x)}{Q(x)} = \frac{P(x_0)}{Q(x_0)}\)

Proof: Algebra of limits \(\lim_{x\to x_0}\frac{P(x)}{Q(x)}=\cfrac{\lim_{x\to x_0}P(x)}{\lim_{x\to x_0}Q(x)}=\cfrac{\lim_{x\to x_0}a_{n}x^{n}+\ldots+\lim_{x\to x_0}a_0}{\lim_{x\to x_0}b_{m}x^{m}+\ldots+\lim_{x\to x_0}b_0}\).

We know that \(\lim_{x \to x_0} c = c\). \(\lim_{x \to x_0} x = x_0\) (here)

Therefore: \(\lim_{x\to x_0}\frac{P(x)}{Q(x)}=\frac{a_{n}x_0^{n}+\ldots+a_0}{b_{m}x_0^{m}+\ldots+b_0}=\frac{P(x_0)}{Q(x_0)}\).

Continuous

Definition

Continuous at point

Let \(f : X \to \mathbb{R}\) and let \(x_0 \in X\).

We say that \(f\) is continuous at \(x_0\) if \(\forall \epsilon > 0, \exists \delta > 0\) such that \(|x-x_0|<\delta\implies|f(x)-f(x_0)|<\epsilon\).

This is the same as saying that: \(\lim_{x \to x_0} f(x) = f(x_0)\) always

If \(x_0\) is a limit point of \(X\) and in \(X\), then \(f\) is continuous at \(x_0 \iff \lim_{x \to x_0} f(x) = f(x_0)\).

Exercise: All rational functions \(\frac{P(x)}{Q(x)}\) are continuous at every \(x_0\) where \(Q(x_0) \neq 0\).

Since \(\lim_{x \to x_0} \frac{P(x)}{Q(x)} = \frac{P(x_0)}{Q(x_0)}\) and when \(x=x_0\), \(\frac{P(x)}{Q(x)}=\frac{P(x_0)}{Q(x_0)}\), then \(\frac{P(x)}{Q(x)}\) are continuous at every \(x_0\) where \(Q(x_0) \neq 0\).

Or prove \(\lim_{x \to x_0} c = c\). \(\lim_{x \to x_0} x = x_0\) is continuous?

Take any \(x_0\), \(\forall\varepsilon,\exists\delta>0,\left|x-x_0\right|<\delta:0<\varepsilon\)

Take any \(x_0\), \(\forall\varepsilon,\exists\varepsilon>0,\left|x-x_0\right|<\varepsilon:\left|x-x_0\right|<\varepsilon\)

Continuous function

If \(f : X \to \mathbb{R}\), we say that \(f\) is continuous on \(X\) if \(f\) is continuous at \(x_0\) for all \(x_0 \in X\).

Exercise: All rational functions are continuous.

Theorem(Algebra of continuous functions)

The sum, the product, and the quotient (on its domain) of continuous functions are continuous.

Proof

Using the limit algebra

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