We will first summarize the basic properties of limits, and then prove each one afterwards.

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

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

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.

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 $\lim\limits_{x\to{c}}k\cdot{f(x)}$:

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

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:

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 $k\ne0$):

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

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

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

This proves the scalar product rule for limits.

Proof. Suppose the following is true:

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

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:

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

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

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

The right-hand side implies the following two inequalities:

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

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:

Let's now add these two inequalities:

From the triangle inequalitylink, we have that:

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

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

This completes the proof.

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:

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

By the triangle inequalitylink, we have that:

Let's summarize the result we have so far:

Therefore, if we can show that:

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:

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

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:

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

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:

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

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:

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:

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

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

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

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

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

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:

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:

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

By definition of limits of functionslink, we conclude that:

This completes the proof.

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:

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

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

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:

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:

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

The second statement is:

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:

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

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

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

Taking the reciprocal gives:

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

To summarize, we have shown that:

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

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:

This completes the proof.

Proof. Our plan of attack is to show that:

We will then be able to conclude our theorem:

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

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:

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:

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:

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

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

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

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

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}:

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

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:

This completes the proof.

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:

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

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

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

This completes the proof.

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:

Using the summation rule of limitslink, we have that:

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

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

Writing the absolute value in terms of interval inequality:

Now, focus on the right inequality:

Adding $M$ and subtracting $L$ gives:

This means that:

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:

This completes the proof.

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:

Because $g(x)$ is zero:

By the rule of limits of constantslink, we get:

This completes the proof.