Mathematical Analysis
Lecture 8
4.3.1 Properties of limits
Corollary 4.3.1. Let $S \subset \R$ and let $c$ be an accumulation point of $S$. Suppose $f \colon S \to \R$ and $g \colon S \to \R$ are functions such that the limits of $f(x)$ and $g(x)$ as $x$ goes to $c$ both exist, and \begin{equation*} f(x) \leq g(x) \qquad \text{for all } x \in S. \end{equation*} Then $$\ds \lim_{x\to c} f(x) \leq \lim_{x\to c} g(x) .$$
4.3.1 Properties of limits
Corollary 4.3.2. Let $S \subset \R$ and let $c$ be an accumulation point of $S$. Suppose $f \colon S \to \R$ is a function such that the limit of $f(x)$ as $x$ goes to $c$ exists. Suppose there are two real numbers $a$ and $b$ such that \begin{equation*} a \leq f(x) \leq b \qquad \text{for all } x \in S. \end{equation*} Then \begin{equation*} a \leq \lim_{x\to c} f(x) \leq b . \end{equation*}
4.3.1 Properties of limits
Corollary 4.3.3. Let $S \subset \R$ and let $c$ be a cluster point of $S$. Suppose $f \colon S \to \R$, $g \colon S \to \R$, and $h \colon S \to \R$ are functions such that
$ f(x) \leq g(x) \leq h(x) \qquad \text{for all } x \in S. $
Suppose the limits of $f(x)$ and $h(x)$ as $x$ goes to $c$ both exist, and
$ \lim_{x\to c} f(x) = \lim_{x\to c} h(x) . $
Then the limit of $g(x)$ as $x$ goes to $c$ exists and
$ \lim_{x\to c} g(x) = \lim_{x\to c} f(x) = \lim_{x\to c} h(x) . $
4.3.1 Properties of limits
Corollary 4.3.4 Let $S \subset \R$ and let $c$ be a cluster point of $S$. Suppose $f \colon S \to \R$ and $g \colon S \to \R$ are functions such that the limits of $f(x)$ and $g(x)$ as $x$ goes to $c$ both exist. Then
- $\displaystyle \lim_{x\to c} \bigl(f(x)+g(x)\bigr) = \lim_{x\to c} f(x) + \lim_{x\to c} g(x). $
- $\displaystyle \lim_{x\to c} \bigl(f(x)-g(x)\bigr) = \lim_{x\to c} f(x) - \lim_{x\to c} g(x). $
- $\displaystyle \lim_{x\to c} \bigl(f(x)\cdot g(x)\bigr) = \lim_{x\to c} f(x)\cdot \lim_{x\to c} g(x). $
- If $\displaystyle \lim_{x\to c} g(x) \not= 0$, and $g(x) \not= 0$ for all $x \in S \setminus \{ c \}$, then \begin{equation*} \lim_{x\to c} \frac{f(x)}{g(x)} = \frac{\lim_{x\to c} f(x)}{\lim_{x\to c} g(x)} . \end{equation*}
4.3.1 Properties of limits
Corollary 4.3.5. Let $S \subset \R$ and let $c$ be an accumulation point of $S$. Suppose $f \colon S \to \R$ is a function such that the limit of $f(x)$ as $x$ goes to $c$ exists. Then \begin{equation*} \lim_{x\to c} \abs{\,f(x)} = \abs{\lim_{x\to c} f(x)}. \end{equation*}
Cluster points?
Why do we need the concept of cluster point in the definition of limit? 🤔
Suppose \(c\) is not a cluster point of \(S\). Then there exists some \(r\) such that the interval \((c-r,c+r)\) contains no points in \(S\) other than \(c\) itself.
Cluster points?
Why do we need the concept of cluster point in the definition of limit? 🤔
Suppose \(c\) is not a cluster point of \(S\). Then there exists some \(r\) such that the interval \((c-r,c+r)\) contains no points in \(S\) other than \(c\) itself.
Then, for any function \(f\), choose any number \(L\).
Let \(\varepsilon\gt0\) be arbitrary and choose \(\delta = r.\) Since there are no points \(x\in S\setminus\{c\}\) such that \(|x-c|\lt\delta=r\), then the statement:
For every \(\varepsilon \gt 0\) there exists a \(\delta \gt 0\) such that if for every \(x\in S\setminus\{c\}\) and \(|x-c|\lt \delta\), then \(|f(x)-L|\lt \varepsilon\).
is vacuously true!
4.4 One-sided limits
Definition 4.4.1. Let $f \colon S \to \R$ be function and let $c$ be a cluster point of $S \cap (c,\infty).$ Then if the limit of the restriction of $f$ to $S \cap (c,\infty)$ as $x \to c$ exists, define \begin{equation*} \lim_{x \to c^+} f(x) := \lim_{x\to c} f\big|_{S \cap (c,\infty)}(x) . \end{equation*} Similarly, if $c$ is a cluster point of $S \cap (-\infty,c)$ and the limit of the restriction as $x \to c$ exists, define \begin{equation*} \lim_{x \to c^-} f(x) := \lim_{x\to c} f\big|_{S \cap (-\infty,c)}(x) . \end{equation*}
4.4 One-sided limits
Definition 4.4.2.
Let $f \colon S \to \R$ be function and let $c$ be a cluster point of
$S \cap (c,\infty)=\{x\in S: x\gt c\}.$
Then we say that $L\in \R$ is a
right-hand limit of $f$ at $c$ and we write
\[
\lim_{x \to c^+} f(x) = L
\]
if given any $\epsilon\gt 0$ there exists a $\delta \gt 0$
such that for all $x\in S$ with $0 \lt x-c \lt \delta,$ then $\abs{\,f(x)-L}\lt
\epsilon.$
4.4 One-sided limits
Definition 4.4.2. Similarly, if $c$ is a cluster point of $S \cap (-\infty,c)=\{x\in S:x\lt c\}.$ Then we say that $L\in \R$ is a left-hand limit of $f$ at $c$ and we write \[ \lim_{x \to c^-} f(x) = L \] if given any $\epsilon\gt 0$ there exists a $\delta \gt0$ such that for all $x\in S$ with $0\lt c-x \lt \delta,$ then $\abs{\,f(x)-L}\lt \epsilon.$
4.4 One-sided limits
4.4 One-sided limits
Right-hand limit
4.4 One-sided limits
Left-hand limit
4.4 One-sided limits
Theorem 4.4.1. Let $S \subset \R$ be such that $c$ is a cluster point of both $S \cap (-\infty,c)$ and $S \cap (c,\infty),$ let $f \colon S \to \R$ be a function, and let $L \in \R.$ Then $c$ is a cluster point of $S$ and \begin{equation*} \lim_{x \to c} f(x) = L \; \iff\; \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L . \end{equation*}
Proof of Theorem 4.4.1
First, we know that $c$ is a cluster point of \[ S,\; S\cap \left(-\infty, c\right)\;\text{ and }\; S\cap \left(c, \infty \right). \]
Also note that \[ S \setminus \{c\}= \big[ S\cap \left(-\infty, c\right)\big] \cup \big[ S\cap \left(c, \infty \right)\big]. \]
Proof of Theorem 4.4.1
$\nec$ Assume $\ds \lim_{x\to c}f(x) = L.$ Given $\vre \gt 0$, there exists $\delta \gt 0$ such that for every $x\in S\setminus \{c\}$ and $\abs{x-c}\lt \delta$, then \[ \abs{f(x)-L}\lt \vre . \qquad (\star) \]
If $x\in S\cap \left(-\infty, c\right)$ $\subset S\setminus \{c\},\,$ then $x\in S\setminus \{c\}$ and $x\lt c \implies 0\lt c-x.$
That is $\,\abs{x-c}\lt \delta\,$ implies $\,0\lt c-x\lt \delta .$ Using $(\star)$ we obtain \[ \abs{f\big|_{S\cap \left(-\infty, c\right)}(x) -L}\lt \vre. \]
Proof of Theorem 4.4.1
$\nec$ Now if $x\in S\cap \left(c,\infty\right)$ $\subset S\setminus \{c\},\,$ then $x\in S\setminus \{c\}$ and $c\lt x \implies 0\lt x-c.$
That is $\,\abs{x-c}\lt \delta\,$ implies $\,0\lt x-c\lt \delta.$ Again, using $(\star)$ we have \[ \abs{f\big|_{S\cap \left(c, \infty\right)}(x) -L}\lt \vre. \]
Therefore
$ \ds \lim_{x\to c^+}f(x) $ $ = L $ $ \ds = \lim_{x\to c^-}f(x). \;\bs $
4.5.1 Limits at infinity
| Limits at infinity | Infinite limits |
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4.5.1 Limits at infinity
We need to define the notion of the limit of a function as $x\to \infty.$ The definition as $x\to -\infty$ is similar.
| Limits at infinity | Infinite limits |
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4.5.1 Limits at infinity
Definition 4.5.2. Let $f \colon (a,\infty) \to \R$ be a function, for some $a\in \R.$ We say that $L\in \R$ is a limit of $f$ as $x\ra \infty,$ if given any $\vre \gt 0,$ there is an $M \gt a$ such that for any $x> M,$ then \begin{equation*} \abs{f(x) - L} \lt \vre. \end{equation*} We write $$ \ds \lim_{x \to \infty} f(x) := L . $$ Alternatively, $f(x) \to L$ as $x \to \infty.$
4.5.1 Limits at infinity
We have, of course, the analogous definition when $x\ra-\infty.$ Let $f \colon (-\infty, a) \to \R.$ We say that $L\in \R$ is a limit of $f$ as $x\ra -\infty,$ if given any $\vre \gt 0,$ there is an $M \lt a$ such that for any $x \lt M,$ then \begin{equation*} \abs{f(x) - L} \lt \vre . \end{equation*} We write \begin{equation*} \lim_{x \to -\infty} f(x) := L , \end{equation*} or alternatively, $f(x) \to L$ as $x \to -\infty .$
Example 4.5.1.
Claim: $\displaystyle \lim_{x\ra \infty} \frac{1}{x} = 0.$
Discussion: Given $\vre \gt 0,$ we are looking for $M$ such that if $x \gt M ,$ then
$\ds\abs{\frac{1}{x} - 0}$ $\ds= \abs{\frac{1}{x} } $ $\ds = \frac{1}{\abs{x}} $ $\lt \vre .$
In order to use the inequality $x \gt M$ we must choose $M\gt 0$ so that
$x \gt M \gt0 $ $\ds\;\implies\; \frac{1}{M} \gt \frac{1}{x} \gt 0.$
Example 4.5.1.
Claim: $\displaystyle \lim_{x\ra \infty} \frac{1}{x} = 0.$
Discussion: Also we need to choose $M$ large enough so that $1/M \lt \vre .$ Then $x \gt M$ implies
$\ds \abs{\frac{1}{x} - 0 }$ $\ds= \frac{1}{\abs{x}}$ $\ds = \frac{1}{x} $ $\lt \ds \frac{1}{M}$ $\lt \vre .$
Now we are ready to write the formal proof.
Example 4.5.1.
Claim: $\displaystyle \lim_{x\ra \infty} \frac{1}{x} = 0.$
Proof. Given an arbitrary $\vre\gt 0$. Choose $M \gt 0 $ large enough so that $1/M \lt \vre .$ For any $x \gt M$, then
$\ds\abs{\frac{1}{x} - 0 } $ $\ds \lt \frac{1}{M}$ $\ds \lt \vre. \;\bs$
Example 4.5.2.
Let $\,f(x) := \dfrac{1}{\abs{x}+1}$. Then
i) $\,\ds \lim_{x\to \infty} f(x) = 0\,$ and ii) $\,\ds\lim_{x\to -\infty} f(x) = 0.$
Proof. Let $\vre \gt 0$ be given. Choose $M \gt 0$ large enough so that $\frac{1}{M+1} \lt \vre .$ If $x \gt M$, then
$\ds\abs{\frac{1}{\abs{x}+1} } $ $\ds = \frac{1}{\abs{x}+1}$ $\ds = \overbrace{ \frac{1}{x+1} }^{ \text{since } x>0}$ $\ds \lt \frac{1}{M+1} $ $\ds \lt \vre.$
So the first limit is proved. 😃
Try to prove the second limit. 📝
4.5.1 Limits at infinity
Theorem 4.5.1. Suppose $f \colon S \to \R$ is a function, $\infty$ is a cluster point of $S \subset \R$, and $L \in \R$. Then \begin{equation*} \lim_{x\to\infty} f(x) = L \end{equation*} if and only if \begin{equation*} \lim_{n\to\infty} f(x_n) = L \end{equation*} for all sequences $\{ x_n \}$ in $S$ such that $\lim\limits_{n\to\infty} x_n = \infty$.
4.5.2 Infinite limit
Definition 4.5.3. Let $f \colon (a,\infty) \to \R$ be a function. We say $f(x)$ diverges to infinity as $x$ goes to $\infty,$ if for every $N \in \R$ there exists an $M\gt a$ such that for all $x \gt M$ then \begin{equation*} f(x) \gt N. \end{equation*} We write \begin{equation*} \lim_{x \to \infty} f(x) := \infty , \end{equation*} or we say that $f(x) \to \infty$ as $x \to \infty.$
Example 4.5.4.
Claim: $\;\displaystyle \lim_{x \to \infty} \frac{1+x^2}{1+x}$ $= \infty.$
Proof. First note that for $x \gt 1,$ we have
$\ds \frac{1+x^2}{1+x} $ $\ds \gt \frac{x^2}{x+x} $ $\ds = \frac{x}{2} .$
Given $N \in \R,$ take $M := \max \{ 2N +1, 1 \}.$ If $x \gt M,$ then $x \gt 1$ and $\dfrac{x}{2} \gt N . $ So
$\ds \frac{1+x^2}{1+x} $ $\ds \gt \frac{x}{2}$ $\ds \gt N . \;\; \bs$