Skip to content

11.18&21 Limit of functions

Limit of functions

Let \(f: X \to \mathbb{R}\), \(X \subseteq \mathbb{R}\). We are led by the following question: If \(a_n \subseteq X\) and \(\lim_{n \to \infty} a_n = \ell\), \(\ell \in X\), is it true that \(\lim_{n \to \infty} f(a_n) = f(\ell)\)?

Yes, when the function is continuous.

Definition

Definition: Let \(f: X \to \mathbb{R}\), \(X \subseteq \mathbb{R}\), and let \(x_0\) be a limit point of \(X\). Then we say \(\lim_{x \to x_0} f(x) = \ell\) if \(\forall \epsilon > 0, \exists \delta > 0\) such that \(0<|x-x_0|<\delta,x\in X\implies|f(x)-\ell|<\epsilon\).

image

Example

  1. Let \(f : \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = x\). Then \(\lim_{x \to x_0} f(x) = x_0\).

    Proof: Given \(\varepsilon > 0\), take \(\delta = \varepsilon\). Now: \(0 < |x - x_0| < \delta \implies |x - x_0| < \varepsilon \implies |f(x) - x_0| < \varepsilon\).

  2. If \(f : \mathbb{R} \to \mathbb{R}\) and \(f(x) = c\), then \(\lim_{x \to x_0} f(x) = c\)​

    Proof: Given \(\varepsilon > 0\), take \(\delta = 3\).

    Now: \(0<|x-x_0|<3\implies.\ldots\implies|f(x)-c|<\varepsilon\).

  3. \(f: \mathbb{R} \to \mathbb{R}\). \(f(x)=\begin{cases}x & x\in\mathbb{Q},\\ 0 & x\notin\mathbb{Q}.\end{cases}\)

    Ex: \(\lim_{x \to 0} f(x) = 0\) \(\lim_{x \to x_0, \, x_0 \neq 0} f(x)\) does not exist.

    image​

  4. Let \(f : \mathbb{R} \to \mathbb{R}\) be defined as follows:

    \(f(x)=\begin{cases}\frac{1}{q}, & \text{if }x=\frac{p}{q}\in\mathbb{Q},\text{ where }\frac{p}{q}\text{ is in reduced form and }q>0,\\ 0, & \text{if }x\notin\mathbb{Q}\text{ or }x=0\end{cases}\)​

    image​

    Find \(\lim_{x \to 1} f(x) = \, ?\)

    Hypothesis: \(|\frac{p}{q}-1|<\delta\) Thesis: \(\frac1q<\varepsilon\)

    Since \(|\frac{p}{q}-1|<\delta\), then \(|\frac{p-q}{q}|<\delta\Rightarrow\frac{1}{q}<\frac{\left|p-q\right|}{q}<\delta\Rightarrow\delta>\frac{1}{q}\)

    Thus we can just choose \(\delta =\varepsilon\)

    Or

    \(\lim_{x \to 1} f(x) = 0\): Let \(\varepsilon > 0\). We need to show that \(|f(x) - 0| < \varepsilon\).

    image​

    Let \(q_0\) be the first such that \(\frac{1}{q_0} < \varepsilon\).

    Let \(A=\left\{\frac11,\frac12,\frac22,\frac13,\frac23,\frac33,\frac14,\frac24,\frac34,\frac44,\dots,\frac{1}{q_0-1},\frac{2}{q_0-1},\frac{3}{q_0-1},\dots,\frac{q_0-1}{q_0-1}\right\}.\)​

    \(A\) is exactly the set of \(a \in \mathbb{R}\) such that \(a > 0\) and \(|f(a)| > \varepsilon\).

    Let \(\delta\) be smaller than \(|a - 1|\) for all \(a \in A, a \neq 1\)

    So that \(0<\delta < |a - 1|\) for all \(a \in A, a \neq 1\)

    Then consider \(0 < |x - 1| < \delta \implies x \notin A \implies |f(x) - 0| < \varepsilon\).

    (\(\delta = \frac{1}{q_0}\) works.)

    This proves that \(\lim_{x \to 1} f(x) = 0\).

    And "the same argument" shows that \(\lim_{x \to x_0} f(x) = 0\) for all \(x_0 \in \mathbb{R}\).

Proposition

Recall

If \(f: X \to \mathbb{R}\) and \(\lim_{x \to x_0} f(x) = \ell_1\) and \(\lim_{x \to x_0} f(x) = \ell_2\), then \(\ell_1 = \ell_2\).

Proof

We have \(\forall\epsilon_1>0,\exists\delta_1>0\) such that \(0<|x-x_0|<\delta_1,x\in X\implies|f(x)-\ell_1|<\epsilon_1\).

Then we have \(\ell_1-\varepsilon_1<f(x)<\varepsilon_1+\ell_1\)​

And \(\forall\epsilon_2>0,\exists\delta_2>0\) such that \(0<|x-x_0|<\delta_2,x\in X\implies|f(x)-\ell_2|<\epsilon_2\).

Then we have \(\ell_2-\varepsilon_2<f(x)<\varepsilon_2+\ell_2\)

Let's take \(\varepsilon=\varepsilon_1=\varepsilon_2=\frac{|\ell_1-\ell_2|}{2}>0\)

Assume \(\ell_1>\ell_2\), then \(\frac{\ell_1+\ell_2}{2}<f(x)\) and \(\frac{\ell_1+\ell_2}{2}>f(x)\)

Contradiction, thus they must be same.

imageimage​

Algebra of Limits

If \(\lim_{x \to a} f(x) = l_1\) and \(\lim_{x \to a} g(x) = l_2\) and \((f, g : X \to \mathbb{R}, a \text{ is a limit point of } X)\)

Then:

  1. \(\lim_{x \to a} (f + g)(x) = l_1 + l_2\)

    \(\forall \epsilon > 0, \exists \delta > 0\) such that \(0<|x-a|<\delta\implies|f(x)-\ell_1|<\frac{\epsilon}{2}\)​

    \(\forall \epsilon > 0, \exists \delta > 0\) such that \(0<|x-a|<\delta\implies|g(x)-\ell_2|<\frac{\epsilon}{2}\)

    \(|f\left(x\right)+g(x)-\left(\ell_1+\ell_2\right)|=|f\left(x\right)-\ell_1+g(x)-\ell_2|\leq|f\left(x\right)-\ell_1\left|+\right|g(x)-\ell_2|=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon\)

  2. \(\lim_{x \to a} (f \cdot g)(x) = l_1 \cdot l_2\)

    \(\lim_{x\to a}f(x)=\ell_1,\,\forall\frac{\varepsilon}{2\left(1+\left|\ell_2\right|\right)}>0,\,\exists\delta_1>0\,\text{such that}\,\forall x\in\text{Dom}f\,\text{and}\,0<|x-a|<\delta_1,\,\text{then}\,|f(x)-\ell_1|<\frac{\varepsilon}{2\left(1+\left|\ell_2\right|\right)}\)

    \(\lim_{x\to a}g(x)=\ell_2,\,\frac{\forall\varepsilon}{2\left(1+\left|\ell_1\right|\right)}>0,\,\exists\delta_2>0\,\text{such that}\,\forall x\in\text{Dom}g\,\text{and}\,0<|x-a|<\delta_2,\,\text{then}\,|g(x)-\ell_2|<\frac{\varepsilon}{2\left(1+\left|\ell_1\right|\right)}\)

    We must show that \(\forall\varepsilon>0,\,\exists\delta>0\,\text{such that}\,\forall x\in\text{Dom}f\cap\text{Dom}g\,\text{and}\,0<|x-a|<\delta,\,\text{we have}\,|f(x)g(x)-\ell_1\ell_2|<\varepsilon.\)

    \(\left|f(x)g(x)-\ell_1\ell_2\right|=\left|f(x)g(x)-\ell_1g(x)+\ell_1g(x)-\ell_1\ell_2\right|\)

    \(=\left|g(x)(f(x)-\ell_1)+\ell_1(g(x)-\ell_2)\right|\) \(\leq\left|g(x)\right|\cdot\left|f(x)-\ell_1\right|+\left|\ell_1\right|\cdot\left|g(x)-\ell_2\right|\)

    I need an upper bound for \(g(x)\): \(\left|g(x)\right|=\left|g(x)+\ell_2-\ell_2\right|\) \(<\left|g(x)-\ell_2\right|+\left|\ell_2\right|\) \(<1+\left|\ell_2\right|\)

    Assume that \(\left|g(x)-\ell_2\right|<1\) and this holds for \(0 < |x-a| < \delta_3\).

    Back to \(\left|f(x)g(x)-\ell_1\ell_2\right|\leq\left|g(x)\right|\cdot\left|f(x)-\ell_1\right|+\left|\ell_1\right|\cdot\left|g(x)-\ell_2\right|\)

    \(<(1+|\ell_2|)|f(x)-\ell_1|+|\ell_1||g(x)-\ell_2|\)

    \(<(1+|\ell_2|)|f(x)-\ell_1|+(|\ell_1|+1)|g(x)-\ell_2|<\varepsilon\)

    \(\delta = \min\{\delta_1, \delta_2, \delta_3\}\)

    With this \(\delta\), the definition holds.

  3. If \(l_1 \neq 0 \implies \exists \delta > 0 : 0 < |x - a| < \delta \implies \frac{|l_1|}{2} < |f(x)| < \frac{3|l_1|}{2}\). Also, \(a\) is a limit point of \(\tilde{X} = \{x \in X : f(x) \neq 0\}\)​

    \(\exists\delta>0\) such that \(0<|x-a|<\delta,x\in X\implies|f(x)-\ell|<\frac{\left|\ell_1\right|}{2}\)​

    Just take \(\varepsilon=\frac{\left|\ell_1\right|}{2}\)​

  4. If \(l_1 \neq 0 \implies \lim_{x \to a} \frac{1}{f(x)} = \frac{1}{l_1}\)​

    \(\forall\epsilon_1>0,\exists\delta_1>0\) such that \(0<|x-a|<\delta_1,x\in X\implies|f(x)-\ell_1|<\epsilon_1\).

    If \(l_1\neq0\implies\exists\delta_2>0:0<|x-a|<\delta_2\implies\frac{|l_1|}{2}<|f(x)|<\frac{3|l_1|}{2}\).

    \(|\frac{1}{f(x)}-\frac{1}{\ell_1}|=\left|\frac{\ell_1-f\left(x\right)}{\ell_1f\left(x\right)}\right|=\frac{\left|\ell_1-f\left(x\right)\right|}{\left|\ell_1\right|\left|f\left(x\right)\right|}\leq\frac{\varepsilon_1}{\frac{\ell_1^2}{2}}=\frac{2\varepsilon_1}{\ell_1^2}=\varepsilon\)

    Just take \(\varepsilon_1=\frac{\varepsilon_{}\ell_1^2}{2}\) and \(\delta=\min\{\delta_1,\delta_2\}\)​

  5. If \(l_2\neq0\implies\lim_{x\to a}\left(\frac{f}{g}\right)\left(x\right)=\frac{l_1}{l_2}\)​

    Just use times

Sandwich Theorem

If \(f_1, g, f_2 : X \to \mathbb{R}\) and \(a\) is a limit point of \(X\), and \(\lim_{x \to a} f_1(x) = l = \lim_{x \to a} f_2(x)\) and \(f_1(x) \leq g(x) \leq f_2(x)\) \(\forall x \in X, x \neq a\), then \(\lim_{x \to a} g(x) = l\).

Proof: Let \(\varepsilon > 0\). We know that \(\exists \delta_1, \delta_2 > 0\) such that: \(\begin{cases} 0<|x-a|<\delta_1\implies|f_1(x)-l|<\varepsilon,\\ 0<|x-a|<\delta_2\implies|f_2(x)-l|<\varepsilon.\end{cases}\)​

Let \(\delta = \min(\delta_1, \delta_2)\).

Now: \(f_1(x)\leq g(x)\leq f_2(x)\implies f_1(x)-l\leq g(x)-l\leq f_2(x)-l\)

\(\implies |g(x) - l| \leq \max\{|f_1(x) - l|, |f_2(x) - l|\}\).

Therefore: If \(0 < |x - a| < \delta\), then \(0 < |x - a| < \delta_1\) and \(0 < |x - a| < \delta_2\), \(\implies\max\{|f_1(x)-l|,|f_2(x)-l|\}<\varepsilon\implies|g(x)-l|<\varepsilon\)

Theorem

Let \(g : X \to \mathbb{R}, f : Y \to \mathbb{R}\). Suppose that \(g(x) \in Y \ \forall x \in X \ (g(X) \subseteq Y)\).

Suppose \(a\) is a limit point of \(X\) and \(\lim_{x \to a} g(x) = b\) and \(b\) is a limit point of \(Y\) and \(\lim_{y \to b} f(y) = c\) where \(g(x)\neq b,\forall x\) (Since \(f\) may not defined at \(b\))

Then \(\lim_{x \to a} f(g(x)) = c\).

Proof: Let \(\varepsilon > 0\).

We know that \(\exists \delta_1 > 0 : 0 < |y - b| < \delta_1 \implies |f(y) - c| < \varepsilon\) (since \(\lim_{y \to b} f(y) = c\)).

Take \(\varepsilon_2=\delta_1>0\)​

Since \(\lim_{x \to a} g(x) = b\), so \(\exists \delta_2 > 0 : 0 < |x - a| < \delta_2 \implies |g(x) - b| < \varepsilon_2 = \delta_1\)

Let \(\delta = \delta_2 > 0\). If \(0 < |x - a| < \delta\), then \(|g(x) - b| < \delta_1\)​\(\implies |f(g(x)) - c| < \varepsilon\).

Proposition

Suppose that \(\forall x\in(x_0-\delta,x_0+\delta),|g(x)|<M\), then \(\lim_{x \to x_0} (x - x_0)g(x) = 0\).

Proof:

Given \(\varepsilon > 0\), let \(\delta = \min\left(\frac{\varepsilon}{M}, \delta_0\right)\).

If \(0 < |x - x_0| < \delta\), then \(|x - x_0| < \delta_0 \implies |g(x)| < M\).

Thus, \(|(x - x_0)g(x)| = |x - x_0| \cdot |g(x)| < |x - x_0| \cdot M\)

Then \(|(x-x_0)g(x)|<\delta\cdot M\)​

Since \(\delta \leq \frac{\varepsilon}{M}\), then \(|(x - x_0)g(x)| < \varepsilon\).

‍