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# Properties of limits

schedule Aug 12, 2023
Last updated
local_offer
Calculus
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We will first summarize the basic properties of limits, and then prove each one afterwards.

Theorem.

# Summary of algebraic properties of limits

Suppose we have the following limits:

$$\lim_{x\to{c}}f(x)=L, \;\;\;\;\;\; \lim_{x\to{c}}g(x)=M$$

Where $L$ and $M$ are finite. The algebraic properties of limits are:

\begin{align*} \color{green}\text{Constants}&: \lim_{x\to{c}}a=a \;\;\;\;\;\;\; \text{where }a\in\mathbb{R}\\\\ \color{green}\text{Scalar multiple}&:\lim_{x\to{c}}k\cdot{f(x)}= k\cdot\lim_{x\to{c}}f(x) =kL \;\;\;\;\;\;\; \text{where }k\in\mathbb{R}\\\\ \color{green}\text{Summation rule}&:\lim_{x\to{c}}[f(x)+g(x)]=\lim_{x\to{c}}f(x)+\lim_{x\to{c}}g(x)=L+M\\\\ \color{green}\text{Product rule}&:\lim_{x\to{c}}[f(x)\cdot{g(x)}]=\lim_{x\to{c}}f(x)\cdot\lim_{x\to{c}}g(x)=LM\\\\ \color{green}\text{Reciprocal rule}&: \lim_{x\to{c}}\frac{1}{f(x)}=\frac{1}{\lim\limits_{x\to{c}}f(x)}=\frac{1}{L}\\\\ \color{green}\text{Quotient rule}&:\lim_{x\to{c}}\frac{f(x)}{g(x)} =\frac{\lim\limits_{x\to{c}}f(x)}{\lim\limits_{x\to{c}}g(x)} =\frac{L}{M} \;\;\;\;\;\;\;\text{where }M\ne0\\\\ \color{green}\text{Composite rule}&: \lim_{x\to{c}}f(g(x))=f\Big(\lim_{x\to{c}}g(x)\Big)\\\\ \color{green}\text{Power rule}&:\lim_{x\to{c}}\Big([f(x)]^p\Big)=\Big(\lim_{x\to{c}}f(x)\Big)^p \;\;\;\;\;\;\;\text{where }p\in\mathbb{R} \end{align*}
Theorem.

# Limits of constants

If $a$ is some constant, then:

$$\lim_{x\to{c}}a=a$$

Proof. Let $a$ and $b$ be some constants. We want to show that given any $\varepsilon\gt0$, there exists $\delta\gt0$ such that:

$$$$\label{eq:dV1AMlpmkzyI8i9liu6} \vert{x-c}\vert\lt\delta \implies \vert{f(x)-a}\vert\lt\varepsilon$$$$

Here, the function we have is $f(x)=a$. Substituting this into the above \eqref{eq:dV1AMlpmkzyI8i9liu6} gives:

$$$$\label{eq:hokO5mn7RPO2jtQxCrC} \vert{x-a}\vert\lt\delta \implies 0\lt\varepsilon$$$$

Because $\varepsilon$ must be greater than zero, $0\lt\varepsilon$ will always be true regardless of what $\delta$ we choose and what the value of $a$ is! This completes the proof.

Theorem.

# Scalar product rule for limits

If $\lim\limits_{x\to{c}}f(x)=L$ for finite $L$ and $k$ is a constant, then:

$$\lim_{x\to{c}}k\cdot{f(x)}= k\cdot{\lim_{x\to{c}}}f(x)=kL$$

Proof. Let $\lim\limits_{x\to{c}}f(x)=L$ for finite $L$ and $k$ be some scalar constant.

Firstly, consider the case when $k=0$. Let's compute $\lim\limits_{x\to{c}}k\cdot{f(x)}$:

$$$$\label{eq:wPohatJRS98RGjeJTCh} \lim_{x\to{c}}0\cdot{f(x)}= \lim_{x\to{c}}0=0$$$$

Note that we used the rule for the limits of constantslink for the second equality. Now, let's compute $k\cdot\lim\limits_{x\to{c}}f(x)$:

$$$$\label{eq:t7F2YpO6CmbWMt6GcSt} 0\cdot\lim_{x\to{c}}f(x)=0$$$$

Therefore, we know that the scalar product rule holds for the case when $k=0$.

We now consider the case when $k\ne0$. Let's take advantage of our assumption that $\lim\limits_{x\to{c}}f(x)=L$. Using the definition of limits of functionslink, we know that for any $\varepsilon\gt0$, there exists some $\delta\gt0$ such that:

$$$$\label{eq:Ohirfma1GP0I0PwVOIB} \vert{x-c}\vert\lt\delta \implies \vert{f(x)-L}\vert\lt\varepsilon$$$$

Remember, $\varepsilon$ is any positive number. This means that multiplying $\varepsilon$ by another positive number is allowed since this will just yield another positive number. Therefore, we are free to multiply $\varepsilon$ by $1/{\vert{k}\vert}$, which is a positive number (and not zero because $k\ne0$):

$$$$\label{eq:aYjT0yQLy9fM0zrUQeI} \vert{x-c}\vert\lt\delta \implies \vert{f(x)-L}\vert\lt\frac{\varepsilon}{\vert{k}\vert}$$$$

Now focus on the right-hand side of \eqref{eq:aYjT0yQLy9fM0zrUQeI}. Multiply both sides by $\vert{k}\vert$ to get:

$$$$\label{eq:Yn2tiGPAXeuZyQzVB3h} \vert{x-c}\vert\lt\delta \implies \vert{k}\vert\vert{f(x)-L}\vert\lt\varepsilon$$$$

Using the basic property of absolute values, we have that:

$$$$\label{eq:nsGJfOVD9WOffMFK2Js} \vert{x-c}\vert\lt\delta \implies \vert{k\cdot{f(x)}-kL}\vert\lt\varepsilon$$$$

By definition of limits of functionslink, we have that $k\cdot{f(x)}$ converges to $kL$ as $x$ approaches $c$, that is:

$$\lim_{x\to{c}}k\cdot{f(x)}$$

This proves the scalar product rule for limits.

Theorem.

# Summation rule for limits

If $\lim\limits_{x\to{c}}f(x)=L$ and $\lim\limits_{x\to{c}}g(x)=M$, then:

$$\lim_{x\to{c}}[f(x)+g(x)]= \lim_{x\to{c}}f(x)+\lim_{x\to{c}}g(x) =L+M$$

Proof. Suppose the following is true:

$$\lim_{x\to{c}}f(x)=L,\;\;\;\;\; \lim_{x\to{c}}g(x)=M$$

By definition of limits of functionslink, given some $\varepsilon\gt0$, there exists some $\delta_1\gt0$ such that:

$$$$\label{eq:hwGXpUnjYThS1z6M818} \vert{x-c}\vert\lt\delta_1 \implies \vert{f(x)-L}\vert\lt\varepsilon$$$$

Remember, $\varepsilon$ is any positive number, which means we are free to multiply $\varepsilon$ by any positive number as this will just create another positive number. Therefore, \eqref{eq:hwGXpUnjYThS1z6M818} can be written as:

$$$$\label{eq:dC8DY2I1d2aVgqZZqAq} \vert{x-c}\vert\lt\delta_1 \implies \vert{f(x)-L}\vert\lt \frac{\varepsilon}{2}$$$$

Similarly, we can write the same expression for function $g$ like so:

$$$$\label{eq:l4P3dfNOHd8RPBr0zDg} \vert{x-c}\vert\lt\delta_2 \implies \vert{g(x)-M}\vert\lt \frac{\varepsilon}{2}$$$$

To prove $\lim\limits_{x\to{c}}[f(x)+g(x)]=L+M$, we must show that there exists some $\delta\gt0$ such that:

$$$$\label{eq:NjDNslJmyi9st1ACJgD} \vert{x-c}\vert\lt\delta \implies \Big\vert[f(x)+g(x)]-(L+M)\Big\vert \lt\varepsilon$$$$

Suppose we let $\delta=\min(\delta_1,\delta_2)$. This implies that:

$$$$\label{eq:BlVpEZPUAu2RRLNj8Jv} \vert{x-c}\vert\lt\delta \;\;\;\Longleftrightarrow\;\;\; \vert{x-c}\vert\lt\min(\delta_1,\delta_2)$$$$

The right-hand side implies the following two inequalities:

\label{eq:snlM4jBdmXNq66A830Z} \begin{aligned}[b] \vert{x-c}\vert\lt\delta_1\\ \vert{x-c}\vert\lt\delta_2\\ \end{aligned}

The reason why the above follows from \eqref{eq:BlVpEZPUAu2RRLNj8Jv} is clear when we consider the following simple scenario:

$$5\lt\min(7,8)$$

Clearly, this means that $5\lt7$ and $5\lt8$.

From \eqref{eq:dC8DY2I1d2aVgqZZqAq} and \eqref{eq:l4P3dfNOHd8RPBr0zDg}, we know that the following two results must be true:

\label{eq:Wbnhte3GlG2hpehzEJ5} \begin{aligned}[b] \vert{f(x)-L}\vert\lt \frac{\varepsilon}{2}\\ \vert{g(x)-M}\vert\lt \frac{\varepsilon}{2} \end{aligned}

Let's now add these two inequalities:

\label{eq:hjv2U3rmYw5FNznoHji} \begin{aligned}[b] \vert{f(x)-L}\vert + \vert{g(x)-M}\vert \lt \varepsilon \end{aligned}

From the triangle inequalitylink, we have that:

$$$$\label{eq:muQTjQBmqJPv6yYeRxf} \vert{f(x)-L+g(x)-M}\vert \le \vert{f(x)-L}\vert + \vert{g(x)-M}\vert$$$$

Combining \eqref{eq:hjv2U3rmYw5FNznoHji} and \eqref{eq:muQTjQBmqJPv6yYeRxf} gives:

$$$$\label{eq:jmXXXICqpln6h6PHsUn} \vert{f(x)-L+g(x)-M}\vert \lt \varepsilon$$$$

Rearranging the terms inside the absolute value gives us the result \eqref{eq:NjDNslJmyi9st1ACJgD} that we are after:

$$$$\label{eq:WiuQMD3mOeIDBf1owGb} \Big\vert[f(x)+g(x)]-(L+M)\Big\vert \lt\varepsilon$$$$

This completes the proof.

Theorem.

# Product rule for limits

If $\lim\limits_{x\to{c}}f(x)=L$ and $\lim\limits_{x\to{c}}g(x)=M$, then:

$$\lim\limits_{x\to{c}}[f(x)\cdot{g(x)}]= \lim\limits_{x\to{c}}f(x)\cdot\lim\limits_{x\to{c}}g(x) =LM$$

Proof. To prove $\lim\limits_{x\to{c}}[f(x) \cdot g(x)]=LM$, we must show that there exists some $\delta\gt0$ such that:

$$$$\label{eq:Jufiha9pbTIqaBFoblq} \vert{x-c}\vert\lt\delta \implies \Big\vert[f(x)\cdot{g(x)}]-LM\Big\vert\lt\varepsilon$$$$

The right-hand side of \eqref{eq:Jufiha9pbTIqaBFoblq} can be written as:

\begin{align*} \vert{f(x)\cdot{g(x)}-LM}\vert&= \vert{f(x)\cdot{g(x)}-L\cdot{g(x)}+L\cdot{g(x)}-LM}\vert\\ &=\vert{g(x)\cdot(f(x)-L)+L(g(x)-M)}\vert \end{align*}

By the triangle inequalitylink, we have that:

\begin{align*} \vert{f(x)\cdot{g(x)}-LM}\vert&= \vert{g(x)\cdot(f(x)-L)+L(g(x)-M)}\vert\\ &\le\vert{g(x)}\vert\cdot\vert{f(x)-L}\vert+\vert{L}\vert \cdot\vert{g(x)-M}\vert\\ &\lt{\color{green}\vert{g(x)}\vert\cdot \vert{f(x)-L}\vert+(\vert{L}\vert + 1)\cdot \vert{g(x)-M}\vert} \end{align*}

Let's summarize the result we have so far:

$$$$\label{eq:iysE1cYqZbGIjTODiZj} \vert{f(x)\cdot{g(x)}-LM}\vert \lt{\color{green}\vert{g(x)}\vert \cdot\vert{f(x)-L}\vert+(\vert{L}\vert + 1) \cdot\vert{g(x)-M}\vert}$$$$

Therefore, if we can show that:

$$\varepsilon= \color{green} \vert{g(x)}\vert\cdot \vert{f(x)-L}\vert+(\vert{L}\vert+1) \cdot\vert{g(x)-M}\vert$$

Then we will have our desired result \eqref{eq:Jufiha9pbTIqaBFoblq}.

Since $\lim\limits_{x\to{c}}g(x)=M$, by definition of limits of functionslink, we know that for any positive $\varepsilon\gt0$, there exists some $\delta_1\gt0$ such that:

$$$$\label{eq:lvMMNwbOtIFGE847mIO} \vert{x-c}\vert\lt\delta_1 \implies \vert{g(x)-M}\vert\lt\varepsilon$$$$

Just as we did in the previous prooflink, we can multiply $\varepsilon$ by any positive value since this will just yield another positive value:

$$$$\label{eq:nI1QZKYmoM9XdrPQBun} \vert{x-c}\vert\lt\delta_1 \implies \vert{g(x)-M}\vert\lt \frac{\varepsilon}{2(\vert{L}\vert+1)}$$$$

Here, we've multiplied $\varepsilon$ by $1/(2(\vert{L}\vert+1)$, which is clearly a positive number. Let's combine the inequalities \eqref{eq:iysE1cYqZbGIjTODiZj} and \eqref{eq:nI1QZKYmoM9XdrPQBun} to get:

\label{eq:wduB64Nv5QtfkzFX6Ns} \begin{aligned}[b] \vert{f(x)\cdot{g(x)}-LM}\vert &\lt{\color{green}\vert{g(x)}\vert\cdot \vert{f(x)-L}\vert+(\vert{L}\vert + 1)\cdot\vert{g(x)-M}\vert}\\ &\lt\vert{g(x)}\vert\cdot\vert{f(x)-L}\vert+(\vert{L}\vert + 1)\Big(\frac{\varepsilon}{2(\vert{L}\vert+1)}\Big)\\ &=\vert{g(x)}\vert\cdot\vert{f(x)-L}\vert+\frac{\varepsilon}{2} \end{aligned}

The summary of the result of \eqref{eq:wduB64Nv5QtfkzFX6Ns} is:

$$$$\label{eq:oTJysgAxFrmK7LyLEYP} \vert{f(x)\cdot{g(x)}-LM}\vert \lt\vert{g(x)}\vert\cdot\vert{f(x)-L}\vert +\frac{\varepsilon}{2}$$$$

Similarly, since $\lim\limits_{x\to{c}}f(x)=L$, by definition of limits of functionslink, we know that for any positive $\varepsilon\gt0$, there exists some $\delta_2\gt0$ such that:

$$$$\label{eq:duACWcsx8bzWkdi9K29} \vert{x-c}\vert\lt\delta_2 \implies \vert{f(x)-L}\vert\lt\varepsilon$$$$

Similar to the case of $g(x)$, we can rewrite \eqref{eq:duACWcsx8bzWkdi9K29} as follows:

$$$$\label{eq:qOXsZeiWYgul8aEQBxZ} \vert{x-c}\vert\lt\delta_2 \implies \vert{f(x)-L}\vert\lt \frac{\varepsilon}{2(\vert{M}\vert+1)}$$$$

Here, we've multiplied $\varepsilon$ by $1/(2(\vert{L}\vert+1)$, which is again a positive number. Let's combine the inequalities \eqref{eq:qOXsZeiWYgul8aEQBxZ} and \eqref{eq:oTJysgAxFrmK7LyLEYP} to get:

\label{eq:IuRTScKApDRVfFSRHqe} \begin{aligned}[b] \vert{f(x)\cdot{g(x)}-LM}\vert &\lt\vert{g(x)}\vert \cdot\vert{f(x)-L}\vert +\frac{\varepsilon}{2}\\ &\lt \vert{g(x)}\vert\cdot \frac{\varepsilon}{2(\vert{M}\vert+1)} +\frac{\varepsilon}{2} \end{aligned}

Next, let's prove that $\vert{g(x)}\vert\le\vert{M}\vert+1$. We start from the left-hand side $\vert{g(x)}\vert$ and use the triangle inequalitylink:

\label{eq:rz6tTtkIqYtwFP25FHA} \begin{aligned}[b] \vert{g(x)}\vert=\vert{g(x)+M-M}\vert \;\;\;\;\;\Longleftrightarrow\;\;\;\;\; \vert{g(x)}\vert\le{\vert{g(x)-M}\vert+\vert{M}\vert}\\ \;\;\;\;\;\Longleftrightarrow\;\;\;\;\; \vert{g(x)}\vert-\vert{M}\vert\le{\vert{g(x)-M}\vert} \end{aligned}

Now, by the definition of limits of functionslink, we know that there exists some $\delta_3\gt0$ such that the following is true:

$$$$\label{eq:JwbU2ncebLythhfutKf} \vert{x-c}\vert\lt\delta_3 \implies \vert{g(x)-M}\vert\lt1$$$$

Combining \eqref{eq:rz6tTtkIqYtwFP25FHA} and \eqref{eq:JwbU2ncebLythhfutKf} gives us the result we want:

$$$$\label{eq:WqdX192NggxtxwxiYN8} \vert{g(x)}\vert-\vert{M}\vert\lt1 \;\;\;\;\;\Longleftrightarrow\;\;\;\;\; \vert{g(x)}\vert\lt\vert{M}\vert+1$$$$

Let's combine the inequalities \eqref{eq:IuRTScKApDRVfFSRHqe} and \eqref{eq:WqdX192NggxtxwxiYN8} to get:

\label{eq:kaLORCAgjWydvbCQRWg} \begin{aligned}[b] \vert{f(x)\cdot{g(x)}-LM}\vert &\lt \vert{g(x)}\vert \cdot\frac{\varepsilon}{2(\vert{M}\vert+1)} +\frac{\varepsilon}{2}\\ &\lt (\vert{M}\vert+1) \cdot\frac{\varepsilon}{2(\vert{M}\vert+1)} +\frac{\varepsilon}{2}\\ &= \frac{\varepsilon}{2} +\frac{\varepsilon}{2}\\ &=\varepsilon \end{aligned}

Let's summarize the result \eqref{eq:kaLORCAgjWydvbCQRWg} in one line:

$$$$\label{eq:LlBWE8y4irfmpdaSrge} \vert{f(x)\cdot{g(x)}-LM}\vert\lt\varepsilon$$$$

We've just shown that \eqref{eq:LlBWE8y4irfmpdaSrge} will be true if the red results below are true:

\label{eq:eQDZ9Ggv25LHwH74Dg6} \begin{aligned}[b] \color{blue}\vert{x-c}\vert\lt\delta_1 &\implies \color{red}\vert{g(x)-M}\vert\lt\varepsilon\\ \color{blue}\vert{x-c}\vert\lt\delta_2 &\implies \color{red}\vert{f(x)-L}\vert\lt\varepsilon \\ \color{blue}\vert{x-c}\vert\lt\delta_3 &\implies \color{red}\vert{g(x)-M}\vert\lt1\\ \end{aligned}

For the red results to hold, the blue results have to be true. This means that we have to choose $\delta$ such that all the blue results are true. Let's set $\delta$ to be defined as follows:

$$$$\label{eq:AENRryf8agF4GEnYRrf} \delta=\min(\delta_1,\delta_2,\delta_3)$$$$

Note that since $\delta_1\gt0$, $\delta_2\gt0$ and $\delta_3\gt0$, we know that $\delta\gt0$. What's amazing about this $\delta$ is that all the blue terms will hold true:

\begin{align*} \vert{x-c}\vert\lt\delta \;\;\;\;\Longleftrightarrow\;\;\;\; \vert{x-c}\vert\lt\min(\delta_1,\delta_2,\delta_3) \;\;\;\; \implies \color{blue}\vert{x-c}\vert\lt\delta_1\\ \vert{x-c}\vert\lt\delta \;\;\;\;\Longleftrightarrow\;\;\;\; \vert{x-c}\vert\lt\min(\delta_1,\delta_2,\delta_3) \;\;\;\; \implies \color{blue}\vert{x-c}\vert\lt\delta_2\\ \vert{x-c}\vert\lt\delta \;\;\;\;\Longleftrightarrow\;\;\;\; \vert{x-c}\vert\lt\min(\delta_1,\delta_2,\delta_3) \;\;\;\; \implies \color{blue}\vert{x-c}\vert\lt\delta_3\\ \end{align*}

Therefore, we have that for any $\varepsilon\gt0$, there exists some $\delta\gt0$ such that:

$$$$\label{eq:oUnN6Fo8rJa7quwcCKK} \vert{x-c}\vert\lt\delta \implies \Big\vert[f(x)+g(x)]-(L+M)\Big\vert \lt\varepsilon$$$$

By definition of limits of functionslink, we conclude that:

$$\lim_{x\to{c}} [f(x)\cdot{g(x)}]=LM$$

This completes the proof.

Theorem.

# Reciprocal rule for limits

If $\lim\limits_{x\to{c}}{f(x)}=L$ and $L\ne0$, then:

$$\lim_{x\to{c}}\frac{1}{f(x)}= \frac{1}{\lim\limits_{x\to{c}}f(x)}= \frac{1}{L}$$

Proof. Assume that $\lim\limits_{x\to{c}}f(x)=L$. We want to show that for any $\varepsilon\gt0$, there exists some $\delta\gt0$ such that:

$$$$\label{eq:lcgSsQKlD8xuva18LEg} \vert{x-c}\vert\lt\delta \implies \Big\vert{\frac{1}{f(x)}-\frac{1}{L}}\Big\vert\lt\varepsilon$$$$

Let's start by rewriting the term within the right absolute value:

\label{eq:g54RPydZeAh27v4Zspj} \begin{aligned}[b] \Big\vert\frac{1}{f(x)}-\frac{1}{L}\Big\vert &= \Big\vert\frac{L-f(x)}{L\cdot{f(x)}}\Big\vert\\ &= \frac{\vert{L-f(x)}\vert}{\vert{L\cdot{f(x)}}\vert}\\ &= \frac{\vert{f(x)-L}\vert}{\vert{L}\vert\cdot\vert{f(x)}\vert}\\ \end{aligned}

Let's substitute \eqref{eq:g54RPydZeAh27v4Zspj} into \eqref{eq:lcgSsQKlD8xuva18LEg} to get:

$$$$\label{eq:KwBcIui3tULK38mlyYe} \vert{x-c}\vert\lt\delta \implies \frac{\vert{f(x)-L}\vert}{\vert{L}\vert\cdot\vert{f(x)}\vert} \lt\varepsilon$$$$

Our goal is to show that \eqref{eq:KwBcIui3tULK38mlyYe} is true.

Now, let's go back to our assumption that $\lim\limits_{x\to{c}}f(x)=L$. From the definition of limits of functionslink, this means that for any $\varepsilon\gt0$, there exists some $\delta$ such that:

$$$$\label{eq:JkfXRLEDyutbqA79gbl} \vert{x-c}\vert\lt\delta \implies \vert{f(x)-L}\vert\lt\varepsilon$$$$

Remember, $\varepsilon$ represents any positive number. This means that as long as we use a positive number in place of $\varepsilon$, \eqref{eq:JkfXRLEDyutbqA79gbl} will still hold. Let's create the following two statements. The first statement is:

$$$$\label{eq:mplTiLTwOkbTT0da2FM} \vert{x-c}\vert\lt\delta_1 \implies \vert{f(x)-L}\vert \lt\frac{\vert{L}\vert}{2}$$$$

Here, $\vert{L}\vert/2$ is positive and non zero because $L\ne0$.

The second statement is:

$$$$\label{eq:tto2OuZRoTEAuVEljpr} \vert{x-c}\vert\lt\delta_2 \implies \vert{f(x)-L}\vert\lt \frac{L^2\varepsilon}{2}$$$$

Again, this is allowed because $L^2\varepsilon/2$ is positive. The reason why we use such arbitrary values will be clear shortly.

Now, let's set $\delta=\min(\delta_1,\delta_2)$. This implies that \eqref{eq:mplTiLTwOkbTT0da2FM} and \eqref{eq:tto2OuZRoTEAuVEljpr} are both true.

Next, by the reverse triangle inequalitylink, we have that:

$$$$\label{eq:jCGF3QWPrdHldApLkQy} \Big\vert \vert{f(x)}\vert- \vert{L}\vert \Big\vert\le \Big\vert f(x)- L \Big\vert$$$$

Combining \eqref{eq:jCGF3QWPrdHldApLkQy} and \eqref{eq:mplTiLTwOkbTT0da2FM} gives:

$$$$\label{eq:jRaPsR8WuC5L83Iaul2} \Big\vert \vert{f(x)}\vert- \vert{L}\vert \Big\vert \lt\frac{\vert{L}\vert}{2}$$$$

By the basic property of absolute values, \eqref{eq:jCGF3QWPrdHldApLkQy} can be written as:

$$$$\label{eq:InUWngvGOTkjMmNv8fF} -\frac{\vert{L}\vert}{2}\lt \vert{f(x)}\vert- \vert{L}\vert \lt\frac{\vert{L}\vert}{2}$$$$

Adding $\vert{L}\vert$ gives:

$$$$\label{eq:hWMEd2f4lpkuLcmXtYP} \frac{\vert{L}\vert}{2}\lt \vert{f(x)}\vert \lt\frac{3\vert{L}\vert}{2}$$$$

Taking the reciprocal gives:

$$$$\label{eq:qzTnXhwl7zYQ5nex94g} \frac{2}{\vert{L}\vert}\gt \frac{1}{\vert{f(x)}\vert}\gt \frac{2}{3\vert{L}\vert}$$$$

Using \eqref{eq:tto2OuZRoTEAuVEljpr} and \eqref{eq:qzTnXhwl7zYQ5nex94g}, we can express \eqref{eq:g54RPydZeAh27v4Zspj} as:

$$$$\label{eq:mmiogpY8uXs0ZphGC74} \frac{\vert{f(x)-L}\vert} {\vert{L}\vert\cdot\vert{f(x)}\vert} \lt \frac{2\vert{f(x)-L}\vert}{\vert{L}\vert\cdot \vert{L}\vert} =\frac{2\vert{f(x)-L}\vert}{L^2} \lt\frac{2}{L^2}\cdot\frac{L^2\varepsilon}{2} =\varepsilon$$$$

To summarize, we have shown that:

$$$$\label{eq:Z0CwqzwKRidwXDVzZ5o} \vert{x-c}\vert\lt\delta \implies \frac{\vert{f(x)-L}\vert} {\vert{L}\vert\cdot\vert{f(x)}\vert} \lt \varepsilon$$$$

This matches \eqref{eq:KwBcIui3tULK38mlyYe} is exactly what we wanted to show. This completes the proof.

Theorem.

# Quotient rule for Limits

If $\lim\limits_{x\to{c}}f(x)=L$ and $\lim\limits_{x\to{c}}g(x)=M$ where $M\ne0$, then:

$$\lim_{x\to{c}}\frac{f(x)}{g(x)} =\frac{\lim\limits_{x\to{c}}f(x)}{\lim\limits_{x\to{c}}g(x)} =\frac{L}{M}$$

Proof. Instead of proving using the formal definition of limits of functions, we can use the product rulelink and reciprocal rulelink for limits that we have already proven:

\begin{align*} \lim_{x\to{c}}\frac{f(x)}{g(x)} &=\lim_{x\to{c}}\Big[f(x)\cdot\frac{1}{g(x)}\Big]\\ &=\lim_{x\to{c}}f(x)\cdot\lim_{x\to{c}}\frac{1}{g(x)}\\ &=\lim_{x\to{c}}f(x)\cdot\frac{1}{\lim\limits_{x\to{c}}g(x)}\\ &=\frac{\lim\limits_{x\to{c}}f(x)}{\lim\limits_{x\to{c}}g(x)}\\ \end{align*}

This completes the proof.

Theorem.

# Limit of a composite function

If $f(u)$ is continuous at $u=L$ and $\lim\limits_{x\to{c}}g(x)=L$, then:

$$\lim_{x\to{c}}f(g(x))= f\Big(\lim_{x\to{c}}g(x)\Big)$$

Proof. Our plan of attack is to show that:

$$$$\label{eq:UTtaG3f6jfQkrPA5oDe} {\color{red}\lim_{x\to{c}}f\big(g(x)\big)=f(L)}, \;\;\;\;\;\;\; {\color{green}f\Big(\lim_{x\to{c}}g(x)\Big)=f(L)}$$$$

We will then be able to conclude our theorem:

$$\lim_{x\to{c}}f\big(g(x)\big)= f\Big(\lim_{x\to{c}}g(x)\Big)$$

To show that the red equality is true, we must show that there exists some $\delta\gt0$ such that:

$$$$\label{eq:lP9HgFfrkcBZc1hZocf} \vert{x-c}\vert\lt\delta \implies \vert{f(g(x))-f(L)}\vert\lt\varepsilon$$$$

Where $\varepsilon$ is any positive number, that is, $\varepsilon\gt0$.

Let's now use what we are given to show \eqref{eq:lP9HgFfrkcBZc1hZocf}. Because $f(u)$ is continuous at $u=L$, we have by the definition of continuitylink, that:

$$\lim_{u\to{L}}f(u)=f(L)$$

In other words, $f(u)$ converges to $f(L)$ as $u$ tends to $L$. By the definition of limitslink, this means that there exists some $\delta_1\gt0$ such that:

$$$$\label{eq:UxzDY0GzBRThKK4zaK0} \vert{u-L}\vert\lt\delta_1 \implies \vert{f(u)-f(L)}\vert\lt\varepsilon$$$$

Remember, $u$ is the input of function $f$. Since we are interested in $f\big(g(x)\big)$, we replace the input $u$ with the output of function $g(x)$, that is $u=g(x)$ like so:

$$$$\label{eq:FH58hwwR52FeOb182PG} \vert{g(x)-L}\vert\lt\delta_1 \implies \vert{f(g(x))-f(L)}\vert\lt\varepsilon$$$$

Now, let's use the other condition - we are given that:

$$$$\label{eq:kQKwJYfQCcDiV68Z9RG} \lim_{x\to{c}}g(x)=L$$$$

Again, by the definition of limitslink, for any $\varepsilon_1>0$, there exists some $\delta\gt0$ such that:

$$$$\label{eq:g9ZWYOPn7MJBqqwVAPa} \vert{x-c}\vert\lt\delta \implies \vert{g(x)-L}\vert\lt\varepsilon_1$$$$

However, $\varepsilon_1$ is any positive number so we can replace $\varepsilon_1$ with $\delta_1$ from earlier:

$$$$\label{eq:rKa3Ew0aDguurB3mR8l} \vert{x-c}\vert\lt\delta \implies \vert{g(x)-L}\vert\lt\delta_1$$$$

Let's now combine \eqref{eq:rKa3Ew0aDguurB3mR8l} and \eqref{eq:FH58hwwR52FeOb182PG} to get:

$$$$\label{eq:Aluz5r0BFPIFVTm9NSf} \vert{x-c}\vert\lt\delta \implies \vert{f(g(x))-f(L)}\vert\lt\varepsilon$$$$

This exactly matches \eqref{eq:lP9HgFfrkcBZc1hZocf}, which is what we wanted to show! This means that the red equality from \eqref{eq:UTtaG3f6jfQkrPA5oDe} holds.

We now need to show green equality from \eqref{eq:UTtaG3f6jfQkrPA5oDe}:

$${\color{green}f\Big(\lim_{x\to{c}}g(x)\Big)=f(L)}$$

Fortunately, this is easy - we are given the following:

$$\lim_{x\to{c}}g(x)=L$$

Taking the function $f$ on both sides gives us the green equality!

Since the green and red equality from \eqref{eq:UTtaG3f6jfQkrPA5oDe} holds, we have our desired result:

$$\lim_{x\to{c}}f(g(x))= f\Big(\lim_{x\to{c}}g(x)\Big)$$

This completes the proof.

Theorem.

# Power rule for limits

If $\lim\limits_{x\to{c}}f(x)$ exists and $p\in\mathbb{R}$, then:

$$\lim_{x\to{c}}\Big([f(x)]^p\Big)= \Big(\lim_{x\to{c}}f(x)\Big)^p$$

Proof. Let's define a new function $g(x)=x^p$ where $p\in\mathbb{R}$. We know that $g(x)$ is a continuous function, which means the composition rule of limits must hold:

$$$$\label{eq:srJlrTvoMrCuPIAzzYN} \lim_{x\to{c}}g(f(x))=g(\lim_{x\to{c}}f(x))$$$$

Let's plug in the definition $g(x)=x^p$ into the left-hand side of \eqref{eq:srJlrTvoMrCuPIAzzYN} to get:

$$$$\label{eq:LG7AKuUYN5JuaTvjdCh} \lim_{x\to{c}}g(f(x)) = \lim_{x\to{c}}\Big([f(x)]^p\Big)$$$$

Now, let's plug in the definition of $g(x)=x^p$ into the right-hand side of \eqref{eq:srJlrTvoMrCuPIAzzYN} to get:

$$$$\label{eq:Ku0C48vrelR0ziHtYK0} g(\lim_{x\to{c}}f(x)) = \Big(\lim_{x\to{c}}f(x)\Big)^p$$$$

Equating \eqref{eq:LG7AKuUYN5JuaTvjdCh} and \eqref{eq:Ku0C48vrelR0ziHtYK0} gives us the power rule of limits:

$$\lim_{x\to{c}}\Big([f(x)]^p\Big) = \Big(\lim_{x\to{c}}f(x)\Big)^p$$

This completes the proof.

Theorem.

# Limit inequality theorem

If $f(x)\le{g(x)}$, $\lim\limits_{x\to{c}}f(x)=L$, and $\lim\limits_{x\to{c}}g(x)=M$, then $L\le{M}$.

Proof. Let $f(x)\le{g(x)}$, $\lim\limits_{x\to{c}}f(x)=L$ and $\lim\limits_{x\to{c}}g(x)=M$. We will prove our theorem by contradiction, that, is, we initially assume that:

$$$$\label{eq:kColHISlxMO1HauiobT} L\gt{M}$$$$

Using the summation rule of limitslink, we have that:

$$M-L=\lim_{x\to{c}}g(x)- \lim_{x\to{c}}f(x)= \lim_{x\to{c}}\Big(g(x)-f(x)\Big)$$

By the definition of limits, we know that for any $\varepsilon\gt0$, there exists some $\delta\gt0$ such that:

$$$$\label{eq:GgOt5HvvlNPGe73tdg3} \vert{x-c}\vert\lt\delta \;\;\;\;\implies\;\;\;\; \vert{g(x)-f(x)-(M-L)}\vert\lt\varepsilon$$$$

Since $\varepsilon$ is any positive number, we can let $\varepsilon=L-M$. Substituting $\varepsilon=L-M$ into \eqref{eq:GgOt5HvvlNPGe73tdg3} gives:

$$$$\label{eq:RblUBp6eZ3Rit5P5sjO} \vert{x-c}\vert\lt\delta \;\;\;\;\implies\;\;\;\; \vert{g(x)-f(x)-M+L}\vert\lt{L-M}$$$$

Writing the absolute value in terms of interval inequality:

$$-L+M\lt{g(x)-f(x)-M+L}\lt{L-M}$$

Now, focus on the right inequality:

$$g(x)-f(x)-M+L\lt{L-M}$$

Adding $M$ and subtracting $L$ gives:

$$g(x)-f(x)\lt{0}$$

This means that:

$$g(x)\lt{f(x)}$$

However, this is a contradiction to our initial hypothesis that $f(x)\le{g(x)}$. Therefore, our assumption \eqref{eq:kColHISlxMO1HauiobT} must be false, that is, the following must be true:

$$L\le{M}$$

This completes the proof.

Theorem.

# Limit of positive and negative functions

If $f(x)\ge0$ for all $x$, then $\lim\limits_{x\to{c}}{f(x)}\ge0$. Conversely, if $f(x)\le0$ for all $x$, then $\lim\limits_{x\to{c}}\le0$.

Proof. We prove the case for $f(x)\ge0$ - the negative case is analogous. Let $f(x)\ge0$ and $g(x)=0$. By the limit inequality theoremlink, since $f(x)\ge{g(x)}$, we have that:

$$\lim_{x\to{c}}f(x)\ge \lim_{x\to{c}}g(x)$$

Because $g(x)$ is zero:

$$\lim_{x\to{c}}f(x)\ge \lim_{x\to{c}}0$$

By the rule of limits of constantslink, we get:

$$\lim_{x\to{c}}f(x)\ge0$$

This completes the proof.

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