15 Limit of Functions and Rules, One-Sided Limits

Lecture from: 17.04.2024 | Video: Video ETHZ

Review: Accumulation Points

We’ve talked about accumulation points (or limit points) of a set $D$. Think of these as points $x_0$ that points in $D$ can “crowd around.” Formally, $x_0$ (which could be a real number, $\infty$, or $-\infty$) is an accumulation point if you can find a sequence of points $(a_n)$ within $D$ (but not $x_0$ itself) that marches steadily towards $x_0$.

Intuition: If $x_0$ is an accumulation point, you can get arbitrarily close to $x_0$ using points from $D$ that are distinct from $x_0$. It’s not just about $x_0$ being in $D$; it’s about $D$ having points “infinitesimally near” $x_0$.

Reformulating Accumulation Points

• For a real number $x_0$: For any tiny distance $\delta >0$ you pick, the “punctured neighborhood” around $x_0$ – that’s the interval $(x_0-\delta ,x_0+\delta )$ with $x_0$ itself removed – must contain some points from $D$. In symbols: $((x_0-\delta ,x_0+\delta )\setminus \{x_0\})\cap D\ne \varnothing$. (This matches Definition 3.10.1 in our script.)

• When $x_0=\infty$: No matter how large a number $c$ you choose, there must be points in $D$ that are even larger. So, $(c,\infty )\cap D\ne \varnothing$.

• When $x_0=-\infty$: Similarly, for any large negative number $-c$, there must be points in $D$ further to the left. So, $(-\infty ,-c)\cap D\ne \varnothing$.

Examples: Seeing Accumulation Points in Action

• Example 1 (from Bsp. 3.10.2): Consider the set $D=\{0\}\cup (1,2)$. This set consists of the single point $0$ and the open interval between $1$ and $2$.

  The accumulation points of $D$, let’s call this set $D^{\prime }$, are all the points in the closed interval $[1,2]$. Why? You can get arbitrarily close to any point in $[1,2]$ (including $1$ and $2$ themselves) using points from the open interval $(1,2)$.

  Important Note: The point $0$ is not an accumulation point of $D$. It’s an “isolated point” – there’s a little bubble around $0$ that contains no other points of $D$.

• Example 2: Take $D=[0,\infty )\setminus \{1\}$. This is the non-negative real line with the single point $1$ removed. The accumulation points $D^{\prime }$ are the entire interval $[0,\infty )$ and $\infty$. You can approach any non-negative real number (including $0$ and $1$) with points from $D$. And, for instance, the sequence $a_n=n+1$ (for $n\ge 1$, so $a_n\ne 1$) is in $D$ and heads off to $\infty$.

Definition: The Limit of a Function

Now, let’s talk about the limit of a function $f:D\to \mathbb{R}$ as $x$ approaches an accumulation point $x_0$ of $D$. We say the limit is $A$ (where $A$ can be a real number, $\infty$, or $-\infty$) if the following holds: no matter how you pick a sequence $(a_n)$ of points in $D$ (excluding $x_0$ itself) that converges to $x_0$, the corresponding sequence of function values $(f(a_n))$ must converge to $A$.

Our shorthand: $A={\lim }_{x\to x_0}f(x)$.

If we can’t find such a consistent $A$ (meaning different paths to $x_0$ give different $f(a_n)$ limits, or $f(a_n)$ doesn’t settle down at all), then we say the limit of $f$ as $x\to x_0$ does not exist. (Though, if $f(x)$ consistently goes to $\infty$ or $-\infty$, we often still say the “limit exists” in that extended sense.)

Examples: Limits in Practice

• Consider $f(x)=x$ for $x\in [0,1)$. As $x$ gets closer and closer to $1$ (from the left, since that’s our domain), $f(x)$ gets closer and closer to $1$. So, ${\lim }_{x\to 1}f(x)=1$.

• Let $g(x)=\{\begin{array}{ll}x& \text{if }0\le x<1\\ x+1& \text{if }x=1\end{array}$, defined on $[0,1]$. To find ${\lim }_{x\to 1}g(x)$, we only care about values of $x$ near $1$ but not equal to $1$. For such $x$ (specifically $x<1$), $g(x)=x$. So, ${\lim }_{x\to 1}g(x)=1$. Notice that the actual value $g(1)=1+1=2$ is completely irrelevant for the limit!

• Now, $h(x)=\{\begin{array}{ll}x& \text{if }0\le x<1\\ x+1& \text{if }1\le x\le 2\end{array}$, defined on $[0,2]$. What is ${\lim }_{x\to 1}h(x)$? If we approach $1$ with $x<1$, $h(x)=x\to 1$. But if we approach $1$ with $x>1$ (and $x\le 2$), $h(x)=x+1\to 1+1=2$. Since we get different results depending on how we approach $1$, the limit ${\lim }_{x\to 1}h(x)$ does not exist. (We’ll soon formalize this idea with one-sided limits.)

Remarks on Limits: Connecting to Epsilon-Delta and Continuity

Let $D\subseteq \mathbb{R}$ and $x_0$ be an accumulation point of $D$.

1. The Epsilon-Delta Viewpoint (for $x_0,A\in \mathbb{R}$)

The limit ${\lim }_{x\to x_0}f(x)=A$ (where $A$ is a real number) means: for any desired level of closeness $ϵ>0$ to $A$, you can find a small enough distance $\delta >0$ around $x_0$ such that for any $x$ in $D$ that’s within this $\delta$-distance of $x_0$ (but not $x_0$ itself), $f(x)$ will be within $ϵ$ of $A$.

Formally: $\forall ϵ>0\exists \delta >0\forall x\in ((x_0-\delta ,x_0+\delta )\setminus \{x_0\})\cap D:|f(x)-A|<ϵ$.

Example: For $D=[0,1)$ and $x_0=1$, the relevant neighborhood is $((1-\delta ,1+\delta )\setminus \{1\})\cap [0,1)=(1-\delta ,1)$.

(This is Definition 3.10.3 in our script.)

The Big Picture: There are $3\times 3=9$ possible scenarios for limits: $x_0$ can be finite, $\infty$, or $-\infty$, and independently, the limit $A$ can also be finite, $\infty$, or $-\infty$. Each has its own precise $ϵ-\delta$ (or similar) formulation.

Example of another formulation: ${\lim }_{x\to x_0}f(x)=-\infty$ (for $x_0\in \mathbb{R}$) means: for any large negative number $-c$ (with $c>0$), you can find a $\delta >0$ such that for all $x\in D$ within $\delta$ of $x_0$ (but not $x_0$), $f(x)$ will be even smaller than $-c$.

That is, $\forall c>0\exists \delta >0\forall x\in ((x_0-\delta ,x_0+\delta )\setminus \{x_0\})\cap D:f(x)<-c$.

Challenge yourself: Try to write down the precise definitions for the other 7 cases!

2. Limits and Continuity

If $x_0$ is actually in the domain $D$ of $f$: The function $f$ is continuous at $x_0$ if and only if two things happen:

1. The limit ${\lim }_{x\to x_0}f(x)$ exists and is a finite real number.
2. This limit is equal to the function’s value at $x_0$, i.e., $f(x_0)={\lim }_{x\to x_0}f(x)$.

Example: Consider $f(x)=\{\begin{array}{ll}0& \text{if }x\in [0,1)\cup (1,2]\\ 1& \text{if }x=1\end{array}$, defined on $[0,2]$. Here, ${\lim }_{x\to 1}f(x)=0$ (because for $x$ near $1$ but not $1$, $f(x)=0$). However, $f(1)=1$. Since $f(1)\ne {\lim }_{x\to 1}f(x)$, the function $f$ is not continuous at $x_0=1$.

What if $x_0$ isn’t in the domain? Continuous Extensions

If $x_0$ is not in $D$, we can’t talk about $f$ being continuous at $x_0$. But, if the limit $L={\lim }_{x\to x_0}f(x)$ does exist (and is finite), we can often “patch the hole” or “extend” the function. Define a new function $\tilde{f}$: $\tilde{f}(x)=\{\begin{array}{ll}f(x)& \text{if }x\in D\\ L& \text{if }x=x_0\end{array}$ This $\tilde{f}$ is now defined at $x_0$, and by construction, it is continuous at $x_0$. We call $\tilde{f}$ the continuous extension of $f$ at $x_0$.

Revisiting an earlier example: $f(x)=x$ on $[0,1)$. We found ${\lim }_{x\to 1}f(x)=1$. The continuous extension to include $x_0=1$ is $\tilde{f}(x)=x$ on $[0,1]$.

3. The Familiar Limit Laws Still Hold

If $f,g:D\to \mathbb{R}$ are functions, and both ${\lim }_{x\to x_0}f(x)$ and ${\lim }_{x\to x_0}g(x)$ exist as finite real numbers, then:

• Sum Rule: ${\lim }_{x\to x_0}(f(x)+g(x))={\lim }_{x\to x_0}f(x)+{\lim }_{x\to x_0}g(x)$
• Product Rule: ${\lim }_{x\to x_0}(f(x)\cdot g(x))=\left({\lim }_{x\to x_0}f(x)\right)\cdot \left({\lim }_{x\to x_0}g(x)\right)$

4. Order Preservation in Limits (Comparison)

If $f(x)\le g(x)$ for all $x$ in $D$ (or at least in a punctured neighborhood of $x_0$ within $D$), and both limits exist, then: ${\lim }_{x\to x_0}f(x)\le {\lim }_{x\to x_0}g(x)$

5. The Squeeze Theorem (or Sandwich Lemma) - A Powerful Tool

Suppose you have three functions $g_1(x)$, $f(x)$, and $g_2(x)$ such that $g_1(x)\le f(x)\le g_2(x)$ for all $x$ in $D$ (or near $x_0$). If the “outer” functions $g_1$ and $g_2$ both approach the same limit $L$ as $x\to x_0$, then the “inner” function $f$ is squeezed between them and must also approach $L$. That is, if ${\lim }_{x\to x_0}{g}_1(x)=L$ and ${\lim }_{x\to x_0}{g}_2(x)=L$, then ${\lim }_{x\to x_0}f(x)=L$.

Example: The Important Limit ${\lim }_{x\to 0}\frac{\sin (x)}{x}$

Let $D=\mathbb{R}\setminus \{0\}$ and $f(x)=\frac{\sin (x)}{x}$. What happens as $x\to 0$?

Claim: ${\lim }_{x\to 0}\frac{\sin (x)}{x}=1$.

Recall from Corollary 3.9.2: For $x\in [0,\sqrt{6}]$, we established the inequality $x-\frac{x^3}{3!}\le \sin (x)\le x$.

For $x\in (0,\sqrt{6}]$ (so $x$ is positive and non-zero), we can divide by $x$: $1-\frac{x^2}{3!}\le \frac{\sin (x)}{x}\le 1$ What about negative $x$? Since $\sin (x)$ is an odd function ($sin(-x)=-\sin (x)$) and $x$ is odd, their ratio $\frac{\sin (x)}{x}$ is an even function: $\frac{\sin (-x)}{-x}=\frac{-\sin (x)}{-x}=\frac{\sin (x)}{x}$. So, the inequality $1-\frac{x^2}{3!}\le \frac{\sin (x)}{x}\le 1$ actually holds for all $x\in [-\sqrt{6},\sqrt{6}]\setminus \{0\}$.

Now, let’s look at the limits of the bounding functions as $x\to 0$:

• ${\lim }_{x\to 0}\left(1-\frac{x^2}{3!}\right)=1-0=1$ (polynomials are continuous).
• ${\lim }_{x\to 0}1=1$.

Since $\frac{\sin (x)}{x}$ is squeezed between two functions that both approach $1$ as $x\to 0$, the Squeeze Theorem tells us: ${\lim }_{x\to 0}\frac{\sin (x)}{x}=1$

A similar argument can be used to show that ${\lim }_{x\to 0}\frac{\cos (x)-1}{x}=0$.

Theorem: Limit of Composite Functions

Let $D,E\subseteq \mathbb{R}$, and $x_0$ be an accumulation point of $D$. Consider a function $f:D\to E$. Suppose the limit $y_0:={\lim }_{x\to x_0}f(x)$ exists, and $y_0$ is in the domain $E$. Now, if we have another function $g:E\to \mathbb{R}$ that is continuous at $y_0$, then we can pass the limit through $g$: ${\lim }_{x\to x_0}g(f(x))=g(y_0)=g\left({\lim }_{x\to x_0}f(x)\right)$

Proof Idea: This follows from the sequential definition of limits and the sequential definition of continuity. If $x_n\to x_0$ (with $x_n\in D\setminus \{x_0\}$), then $f(x_n)\to y_0$. Since $g$ is continuous at $y_0$, $g(f(x_n))\to g(y_0)$.

Example: Applying the Composite Limit Theorem

Let’s find ${\lim }_{x\to 0}\frac{1}{1+\frac{\sin (x)}{x}}$. Define $f(x)=\frac{\sin (x)}{x}$. We just showed $y_0={\lim }_{x\to 0}f(x)=1$. Define $g(y)=\frac{1}{1+y}$. This function $g$ is continuous at $y_0=1$ (since $1+1\ne 0$). Therefore, ${\lim }_{x\to 0}\frac{1}{1+\frac{\sin (x)}{x}}=g\left({\lim }_{x\to 0}\frac{\sin (x)}{x}\right)=g(1)=\frac{1}{1+1}=\frac{1}{2}$

One-Sided Limits: Approaching from Left or Right

Observation: Consider $f(x)=\frac{1}{x}$. As $x$ approaches $0$, the behavior is drastically different depending on whether $x$ is positive (approaching from the right) or negative (approaching from the left).

Definition: One-Sided Accumulation Points

Let $D\subseteq \mathbb{R}$.

• $x_0\in \mathbb{R}$ is a right-sided accumulation point of $D$ if $x_0$ is an accumulation point of the set $D\cap (x_0,\infty )$ (points in $D$ strictly to the right of $x_0$).
• $x_0\in \mathbb{R}$ is a left-sided accumulation point of $D$ if $x_0$ is an accumulation point of $D\cap (-\infty ,x_0)$ (points in $D$ strictly to the left of $x_0$).

Definition: One-Sided Limits

Let $D\subseteq \mathbb{R}$, and $A$ be a real number, $\infty$, or $-\infty$.

• If $x_0\in \mathbb{R}$ is a right-sided accumulation point of $D$, the right-sided limit of $f$ as $x$ approaches $x_0$ is $A$, written $A={\lim }_{x\to x_0^{+}}f(x)$, if $A$ is the limit of the function $f$ restricted to $D\cap (x_0,\infty )$ as $x\to x_0$.

• If $x_0\in \mathbb{R}$ is a left-sided accumulation point of $D$, the left-sided limit of $f$ as $x$ approaches $x_0$ is $A$, written $A={\lim }_{x\to x_0^{-}}f(x)$, if $A$ is the limit of the function $f$ restricted to $D\cap (-\infty ,x_0)$ as $x\to x_0$.

Sequential View: ${\lim }_{x\to x_0^{+}}f(x)=A⟺$ for every sequence $(a_n)$ in $D$ with $a_n>x_0$ for all $n$ and $\lim a_n=x_0$, we have $\lim f(a_n)=A$. (And analogously for the left-sided limit with $a_n<x_0$.)

Example: For $f(x)=1/x$: ${\lim }_{x\to 0^{+}}\frac{1}{x}=\infty$ ${\lim }_{x\to 0^{-}}\frac{1}{x}=-\infty$

Example: $\ln (x)$ as $x\to 0^{+}$

Consider $\ln :(0,\infty )\to \mathbb{R}$. We claim ${\lim }_{x\to 0^{+}}\ln (x)=-\infty$.

Proof

Let $(a_n)$ be any sequence in $(0,\infty )$ such that $a_n\to 0$ as $n\to \infty$. We want to show that $\ln (a_n)\to -\infty$.

Pick any large positive number $C$. We want to show $\ln (a_n)<-C$ for $n$ large enough.

Consider $e^{-C}$. Since $0<x<e^{-C}$ implies $\ln (x)<\ln (e^{-C})=-C$ (because $\ln$ is strictly increasing).

Since $a_n\to 0$ and $e^{-C}>0$, there must be an $N\in \mathbb{N}$ such that for all $n\ge N$, we have $0<a_n<e^{-C}$.

For these $n$, it follows that $\ln (a_n)<-C$. Since $C$ was arbitrary, this means $\ln (a_n)\to -\infty$. Thus, ${\lim }_{x\to 0^{+}}\ln (x)=-\infty$. $\square$

Example: $x^a$ as $x\to 0^{+}$ for $a>0$

Consider $f(x)=x^a=\exp (a\ln (x))$ for $a>0$, defined on $(0,\infty )$. Then ${\lim }_{x\to 0^{+}}{x}^a=0$.

Reasoning: As $x\to 0^{+}$, $\ln (x)\to -\infty$. Since $a>0$, $a\ln (x)\to -\infty$. Then $x^a=\exp (a\ln (x))\to \exp (-\infty )$, which we interpret as $0$.

(Using the composite limit theorem: ${\lim }_{y\to -\infty }\exp (y)=0$).

Continuous Extension of $x^a$

For $a>0$, the function $x\mapsto x^a$ on $(0,\infty )$ has a limit of $0$ as $x\to 0^{+}$. So, we can define a continuous extension to include $x=0$:

This extended function is strictly monotonically increasing, bijective (if we consider the codomain $[0,\infty )$), and continuous on its entire domain $[0,\infty )$.

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