The proofs of the generic Limit Laws depend on the definition of the limit. Therefore, we first recall the definition.

$\lim\limits_{x\to c} f(x)=L$ means that for every $\epsilon>0$, there exists a $\delta>0$, such that for every $x$, the expression $0< |x-c|<\delta$ implies $|f(x)-L|<\epsilon$. |

As we prove each rule (in the left-hand column of each table), we shall also provide a running commentary (in the right hand column).

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

Suppose $\epsilon >0$ has been provided. | This is the first line of any delta-epsilon proof, since the definition of the limit requires that the argument work for any epsilon. |

Define $\epsilon_2 = \dfrac{\epsilon}{2}$. | We are defining a new, smaller epsilon. Looking ahead, we see that two functions will be contributing to the variation in the combined sum, therefore we have decided to limit the variation in each function to half of the allowed epsilon variation. |

Then there exists a $\delta_1 >0$ such that for every $x$, the expression $0 < |x-c| <\delta_1$ implies $|f(x)-L| <\epsilon_2$. | Since limits work for any epsilon, they will work for our new epsilon (that is, $\epsilon_2$). So we have used the definition of the given limit $\lim\limits_{x\to c} f(x)=L$ to obtain a delta (specifically $\delta_1$) for that function. |

And there exists a $\delta_2 >0$ such that for every $x$, the expression $0 < |x-c| <\delta_2$ implies $|g(x)-M| <\epsilon_2$. | And we use the definition of the given limit $\lim\limits_{x\to c} g(x)=M$ with our new epsilon to obtain another delta, $\delta_2$, which may in fact have a different value than $\delta_1$. |

Now define $\delta=\min\{\delta_1,\delta_2\}$. Note that $\delta>0$. | This unsubscripted delta is a third delta, but is the delta that is necessary for the completion of our proof. Since it is the minimum of two deltas already known to be positive, it is also positive. |

Then for every $x$, the expression $0 < |x-c| <\delta$ implies | This is wording from the definition of the limit. |

$0 < |x-c| < \min\{\delta_1,\delta_2\}$ $0 < |x-c| <\delta_1$ and $0 < |x-c| <\delta_2$ |
We replaced $\delta$ with the definition we provided for it. Then since the absolute value quantity is less than the minimum of two values, it is also less than each of those quantities separately. |

$|f(x)-L|<\epsilon_2$ and $|g(x)-M|<\epsilon_2$ | Note that these are the conclusions of two of our previous statements. All of the conditions for each of these statements have been satisfied, so both of them are true simultaneously. |

$|f(x)-L|+|g(x)-M|<\epsilon_2+\epsilon_2$ | Here, we have added the two inequalities together. |

But by the Triangle Inequality, we also have $|f(x)-L+g(x)-M| \le |f(x)-L|+|g(x)-M|$ |
The Triangle Inequality states that for any real number values of $A$ and $B$, the inequality $|A+B| \le |A|+|B|$ will be true. |

That gives $|f(x)-L+g(x)-M| <\epsilon_2+\epsilon_2$ | This follows because of the transitivity of inequalities. In other words, if $A < B$ and $B < C$, then $A < C$. |

$|f(x)+g(x)-L-M| <\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}$ $|(f(x)+g(x))-(L+M)| <\epsilon$ |
Then we recall our definition of $\epsilon_2$, perform a little addition, and rearrange terms inside the absolute values. |

Therefore, $\lim\limits_{x\to c} [f(x)+g(x)]=L+M$. | The previous line was the result we needed to substantiate this claim for any two functions. |

- If $\lim\limits_{x\to c} f(x)=L$, then $\lim\limits_{x\to c} [kf(x)]=kL$.

Suppose $\epsilon >0$ has been provided. | This is the first line of any delta-epsilon proof, since the definition of the limit requires that the argument work for any epsilon. |

Since $\lim\limits_{x\to c} 0=0$, the statement is true for $k=0$. Now assume $k \ne 0$. | We first dispense with the trivial case of $k=0$, especially since we need a non-zero value for $k$ to do the remainder of the proof. |

Define $\epsilon_2=\dfrac{\epsilon}{|k|}$. | Here, we define a new epsilon. This new epsilon will still be positive, but may be smaller or larger, depending on the (non-zero) value of $k$. |

Then there exists a $\delta >0$ such that for every $x$, the expression $0 < |x-c| <\delta$ implies $|f(x)-L| <\epsilon_2$. | Since limits work for any epsilon, they will work for our new epsilon (that is, $\epsilon_2$). So we have used the definition of the given limit $\lim\limits_{x\to c} f(x)=L$ to obtain a delta for that function. |

Then $|f(x)-L| < \dfrac{\epsilon}{|k|}$. | Here, we replaced $\epsilon_2$ by its definition. |

$|k(f(x)-L)|<\epsilon$ $|kf(x)-kL|<\epsilon$ |
Then we multiply both sides of the inequality by $|k|$, and distribute. |

Therefore, $\lim\limits_{x\to c} [kf(x)]=kL$. | And the previous line implies the limit statement we wanted to prove. |

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

$\lim\limits_{x\to c} [f(x)-g(x)]$ $=\lim\limits_{x\to c} [f(x)+(-1)g(x)]$ |
Rather than using delta and epsilon, we can use the previous limit rules that we have proven. So we rewrite the difference. |

$=\lim\limits_{x\to c} [f(x)]+\lim\limits_{x\to c} [(-1)g(x)]$ | We can write the expression above as the sum of two limits, because of the Sum Law proven above. |

$=\lim\limits_{x\to c} f(x)+(-1)\lim\limits_{x\to c} g(x)$ | Then we rewrite the second term using the Scalar Multiple Law, proven above. |

$=L+(-1)M$ $=L-M$ |
The values of these two limits were already given in the hypothesis of the theorem. Then algebra finishes the computation. |

Therefore, $\lim\limits_{x\to c} [f(x)-g(x)]=L-M$. | And thus we have our result. |

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

Suppose $\epsilon>0$ has been provided. Then | The necessary first line of any delta-epsilon proof. |

$f(x)g(x)-LM$ $=f(x)g(x)-LM+Lg(x)-Lg(x)+Mf(x)-Mf(x)+LM-LM$ $=Lg(x)-LM+Mf(x)-ML+f(x)g(x)-Mf(x)-Lg(x)+LM$ $=L(g(x)-M)+M(f(x)-L)+(f(x)-L)(g(x)-M)$ |
This preliminary work is needed to separate the functions $f(x)$ and $g(x)$ into pieces whose limits can be separately handled. So we added and subtracted a number of terms, then strategically rearranged and factored them. The result suggests that we need four subsidiary epsilons, since the functions appear in four locations. However, two of the appearances are similar, so we will settle for three epsilons. |

Define the following: $\epsilon_2=\sqrt{\dfrac{\epsilon}{3}}$ $\epsilon_3=\dfrac{\epsilon}{3(1+|M|)}$ $\epsilon_4=\dfrac{\epsilon}{3(1+|L|)}$ |
We are defining three new epsilons in such a way as to keep all of the various parts of our work from combining to cause more variation than the original epsilon allowed. Notice that all of the new epsilons will be positive, since the original epsilon was positive. The rationale for their values comes from the previous preliminary work. That work had three terms in its last line, so we are hoping to restrict the variation in each to one-third of the original epsilon. The fourth epsilon will help control the first term, the third epsilon will control the second term, and the second epsilon will control each of the factors in the last term (hence the square root). The "one plus" quantities in the last two epsilons are present to avoid difficulties if $L$ or $M$ is zero. |

Then there exists a $\delta_1 >0$ such that for every $x$, the expression $0 < |x-c| <\delta_1$ implies $|f(x)-L| <\epsilon_2$. | This is the result of the limit statement $\lim\limits_{x\to c} f(x)=L$ used with $\epsilon_2$. |

And there exists a $\delta_2 >0$ such that for every $x$, the expression $0 < |x-c| <\delta_2$ implies $|g(x)-M| <\epsilon_2$. | This is the result of the limit statement $\lim\limits_{x\to c} g(x)=M$ also used with $\epsilon_2$. |

And there exists a $\delta_3 >0$ such that for every $x$, the expression $0 < |x-c| <\delta_3$ implies $|f(x)-L| <\epsilon_3$. | This is the result of the limit statement $\lim\limits_{x\to c} f(x)=L$ used with $\epsilon_3$. |

And there exists a $\delta_4 >0$ such that for every $x$, the expression $0 < |x-c| <\delta_4$ implies $|g(x)-M| <\epsilon_4$. | This is the result of the limit statement $\lim\limits_{x\to c} g(x)=M$ used with $\epsilon_4$. |

Define $\delta=\min\{\delta_1,\delta_2,\delta_3,\delta_4\}$. Note that $\delta>0$. | We want the result to be true for the smallest possible delta needed. Since all of these deltas were positive, our unsubscripted delta is also positive. |

Then for every $x$, the expression $0 < |x-c| <\delta$ implies | The wording from the limit definition that begins the chain of implications. |

$0 < |x-c| <\delta_1$, $0 < |x-c| <\delta_2$, $0 < |x-c| <\delta_3$, and $0 < |x-c| <\delta_4$ | Since $\delta$ was the minimum of the four subscripted deltas, the previous inequality implies all four of these inequalities. |

which implies $|f(x)-L|<\epsilon_2$, $|g(x)-M|<\epsilon_2$, $|f(x)-L|<\epsilon_3$, and $|g(x)-M|<\epsilon_4$. | These are the four conclusions that result from the existence of the four deltas. |

Now since $|f(x)g(x)-LM|=|L(g(x)-M)+M(f(x)-L)+(f(x)-L)(g(x)-M)|$ | Here, we recall a previous equation above, and apply absolute values to both sides. |

then the Triangle Inequality will give $|f(x)g(x)-LM| \le |L||g(x)-M|+|M||f(x)-L|+|f(x)-L||g(x)-M|$ | The Triangle Inequality is typically written as $|A+B| \le |A|+|B|$, but it can also be generalized to more than two terms. We have also written the absolute value of the products as the product of absolute values. |

which gives $|f(x)g(x)-LM|< |L|\epsilon_4 + |M|\epsilon_3+\epsilon_2\epsilon_2$. | As we replace each absolute value quantity by one of the epsilons obtained above, the strict inequality in the above statements allows us to use a strict inequality in this statement. |

$|f(x)g(x)-LM|< \dfrac{|L|\epsilon}{3(1+|L|)}+\dfrac{|M|\epsilon}{3(1+|M|)}+\sqrt{\dfrac{\epsilon}{3}}\sqrt{\dfrac{\epsilon}{3}}$ | Then we replace each subscripted epsilon with the formula we chose above. |

Now we also know that $0\le\dfrac{|L|}{1+|L|}<1$, and that $0\le\dfrac{|M|}{1+|M|}<1$. This gives | Since both numerator and denominator are always positive, and the denominator is always larger than the numerator, these fractions will always be less than 1. |

$|f(x)g(x)-LM|<\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}$ $|f(x)g(x)-LM|<\epsilon$ |
Here, we replaced the fractional quantities involving $|L|$ and $|M|$ by 1, which does not harm the inequality. |

Therefore, $\lim\limits_{x\to c} [f(x)g(x)]=LM$. | And the previous line confirms this limit statement. |

- If $\lim\limits_{x\to c} g(x)=M$ and $M\ne 0$, then $\lim\limits_{x\to c} \dfrac{1}{g(x)}=\dfrac{1}{M}$.

Suppose $\epsilon>0$ has been provided. | This is how every delta-epsilon proof must begin, because the result must be proved true for every possible positive epsilon. |

Note that $\dfrac{1}{g(x)}-\dfrac{1}{M}=\dfrac{1}{M}\dfrac{1}{g(x)}(M-g(x))$. | This result can be verified by getting a common denominator for the two fractions on the left, and combining the fractions. It suggests that we will need two subsidiary epsilons, since the function appears twice. |

Define $\epsilon_2=\dfrac{|M|}{2}$ and $\epsilon_3=\dfrac{M^2\epsilon}{2}$. | Both of these epsilons will be positive. Comparing these definitions with the preliminary work above, you should notice that when the second factor is replaced by the reciprocal of $\epsilon_2$, and the third factor by $\epsilon_3$, cancellation will occur that will leave only the original epsilon. |

Then there exists a $\delta_3 >0$ such that for every $x$, the expression $0 < |x-c| < \delta_3$ implies $|M-g(x)| <\epsilon_3$. | This is the result of the limit statement $\lim\limits_{x\to c}g(x)=M$. We are also using the fact that $|A-B|=|B-A|$, that is, that the absolute values of opposites are equal. |

And there exists a $\delta_2 >0$ such that for every $x$, the expression $0 < |x-c| < \delta_2$ implies $|g(x)-M| <\epsilon_2$. | This is a straightforward result of the limit statement $\lim\limits_{x\to c} g(x)=M$. |

Define $\delta=\min\{\delta_2,\delta_3\}$. Note that $\delta>0$. Then for every $x$, the expression $0 < |x-c| < \delta$ implies both $0 < |x-c| < \delta_2$ and $0 < |x-c| < \delta_3$. | We choose the minimum of the two deltas, so both restrictions will be met. |

Now by the Triangle Inequality, $|g(x)|-|M|\le |g(x)-M|$, and $|M|-|g(x)|\le |M-g(x)|$, | This is really an immediate corollary of the Triangle Inequality. Specifically, $|g(x)|=|g(x)-M+M|\le |g(x)-M|+|M|$, then $|M|$ can be subtracted from both sides. The second result is similar. Note that the right-hand sides of the two results are equal, since they are absolute values of opposites. And the left-hand sides of the two results are opposites. |

which implies $||g(x)|-|M||\le |g(x)-M|$, $||g(x)|-|M||<\epsilon_2$, $||g(x)|-|M||<\dfrac{|M|}{2}$, |
Since both of the opposite quantities in the previous line were less than the right hand side, then the absolute value of the difference is also less than the right hand side. But since the restriction on $\delta_2$ was met, the inequality is bounded by $\epsilon_2$, which was defined above. |

which further implies $-\dfrac{|M|}{2}<|g(x)|-|M|<\dfrac{|M|}{2}$, $\dfrac{|M|}{2}<|g(x)|<\dfrac{3|M|}{2}$, |
The absolute value inequality in the previous line has been expanded, then we added $|M|$ to each expression. |

and thus $\dfrac{1}{|g(x)|}<\dfrac{2}{|M|}$. | We obtain this by using the left two expressions of the previous inequality, and taking reciprocals of both sides. (Alternatively, divide each side by both numerators, and multiply by 2.) |

Since $\left|\dfrac{1}{g(x)}-\dfrac{1}{M}\right|=\left|\dfrac{1}{M}\dfrac{1}{g(x)}(M-g(x))\right|=\dfrac{1}{|M|}\dfrac{1}{|g(x)|}|M-g(x)|$, | Here we are taking absolute values of a result early in this proof. |

we now have $\left|\dfrac{1}{g(x)}-\dfrac{1}{M}\right| < \dfrac{1}{|M|}\dfrac{2}{|M|}\epsilon_3 < \dfrac{1}{|M|}\dfrac{2}{|M|}\dfrac{M^2\epsilon}{2}=\epsilon$ | We can now bound the second and third factors of the previous line by the restrictions we have found. Then replacing $\epsilon_3$ by the quantity from its definition, we find that the right hand side is simply equal to $\epsilon$. (The leftmost expression is still less than $\epsilon$.) |

Therefore, $\lim\limits_{x\to c} \dfrac{1}{g(x)}=\dfrac{1}{M}$. | Thus, we have reached the goal needed to confirm the limit statement. |

- If $\lim\limits_{x\to c} f(x)=L$ and $\lim\limits_{x\to c} g(x)=M$ and $M\ne 0$, then $\lim\limits_{x\to c} \dfrac{f(x)}{g(x)}=\dfrac{L}{M}$.

$\lim\limits_{x\to c} \dfrac{f(x)}{g(x)}=\lim\limits_{x\to c}\left[f(x)\dfrac{1}{g(x)}\right]$ $=\left(\lim\limits_{x\to c} f(x)\right)\left(\lim\limits_{x\to c} \dfrac{1}{g(x)}\right)$ $=L\left(\dfrac{1}{M}\right)=\dfrac{L}{M}$ |
We can use the previous Limit Laws to prove this rule. First, we rewrite the quotient as a product. Then, we can use the Product Law, followed by the Reciprocal Law. |

Therefore, $\lim\limits_{x\to c} \dfrac{f(x)}{g(x)}=\dfrac{L}{M}$. | And we have the result. |

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

Suppose $\epsilon>0$ has been provided. | As in all delta-epsilon proofs, we begin with an arbitrary positive value for epsilon. |

Then there exists a $\delta_1>0$ such that for every $u$, the expression $0< |u-M|<\delta_1$ implies $|f(u)-f(M)|<\epsilon$. | This is the result of the limit statement $\lim\limits_{u\to M} f(u)=f(M)$. |

But when $u=M$, we also have $f(u)=f(M)$, and $|f(u)-f(M)|=0$. | Since $f(x)$ is a function, its values are unique. |

Therefore there exists a $\delta_1>0$ such that for every $u$, the expression $0\le |u-M|<\delta_1$ implies $|f(u)-f(M)|<\epsilon$. | In order for this proof to work, we need to remove the non-zero restriction on the expression $|u-M|$. The expression $\lim\limits_{u\to M}f(u)$ was not simply equal to an arbitrary value $L$, but to the function value $f(M)$. When a function has this property, it is called a "continuous" function. |

Define $\epsilon_2=\delta_1$. | Since the composition of two functions takes the output of the first as the input of the second, we need a similar result with our deltas and epsilons. Note that this epsilon is positive. |

Then there exists a $\delta>0$ such that for every $x$, the expression $0< |x-c| <\delta$ implies $|g(x)-M|<\epsilon_2$. | This is the result of the limit statement $\lim\limits_{x\to c} g(x)=M$, applied with $\epsilon_2$. The delta from this limit is the delta we need for the composite function. |

Now define $u=g(x)$. | Setting these two quantities equal allows us to use $g(x)$ as the input of $f(x)$. |

Then for every $x$, the expression $0< |x-c| <\delta$ implies | The statement that begins the chain of implications. |

$|g(x)-M|<\epsilon_2$, $|u-M|<\delta_1$, |
The first inequality is the conclusion of a previous statement. The second inequality comes from replacing $u$ by its equivalent $g(x)$, and $\epsilon_2$ by its equivalent $\delta_1$. |

$|f(u)-f(M)|<\epsilon$, and finally $|f(g(x))-f(M)|<\epsilon$. |
The first inequality here is the conclusion of an earlier statement, and we should note that the removal of the non-zero restriction on the previous inequality was necessary to reach this step. The second inequality arises from the first by replacing $u$ with its equivalent $g(x)$. |

Therefore, $\lim\limits_{x\to c} f(g(x))=f(M)$. | And we have proven our result. |

- Suppose $\lim\limits_{x\to c} f(x)=L$, where $a< c< b$. Also suppose that $f(x)$ is defined and strictly monotonic on the set $S=[a,c)\cup (c,b]$, that $\lim\limits_{x\to k} f(x)=f(k)$ for all values $k$ in set $S$ (but using appropriate one-sided limits at values $a$ and $b$), that $g(x)=f(x)$ on set $S$, and that $g(c)$ is undefined. Then $\lim\limits_{u\to L} g^{-1}(u)=c$.

Suppose $\epsilon>0$ has been provided. | This is the original epsilon by which we will bound the variation in the inverse function. |

Define $\epsilon_1=\min\{\epsilon,c-a,b-c\}$ | If the $\epsilon$ given was too large, it would be possible for some later functions to be undefined. Therefore, if needed, we choose a smaller epsilon. Note that all of the candidates for $\epsilon_1$ are still positive. |

and define $\epsilon_2=\min\{|f(c-\epsilon_1)-L|,|f(c+\epsilon_1)-L|\}$. | Both functional values are well defined, and because of the monotonicity of $f(x)$, both expressions are positive (and not zero). |

Then there exists a $\delta_1>0$ such that for every $x$, the expression $0< |x-c|<\delta_1$ implies $|f(x)-L|<\epsilon_2$. | This is the implication of the statement $\lim\limits_{x\to c} f(x)=L$, when applied to the value $\epsilon_2$. |

Now let $u=g(x)$, and $\delta=\epsilon_2$. | The use of the function $g(x)$ rather than $f(x)$ is necessary, because $f(c)$ may be defined in such a way that its inverse would not be a function. But $g(x)$ will have the same monotonicity without having a possible impediment to an inverse function on set $S$. |

Then for all $u$, the expression $0< |u-L|<\delta$ implies $0< |u-L|<\epsilon_2$, which also implies $-\epsilon_2 < u-L<\epsilon_2$. | Here, we begin our chain of implications. By definition above, we had $\epsilon_2=\delta$. |

If $f(x)$ is an increasing function, then we have $f(c-\epsilon_1)-L < u-L < f(c+\epsilon_1)-L$. But if $f(x)$ is a decreasing function, then we have $f(c+\epsilon_1)-L < u-L < f(c-\epsilon_1)-L$. | Then we replace $\epsilon_2$ by appropriate values from its definition. We have to make a distinction between strictly increasing and strictly decreasing functions, so as to keep the functional values on the correct sides of the inequality. |

Then $f(c-\epsilon_1) < u < f(c+\epsilon_1)$ for increasing functions, or $f(c+\epsilon_1) < u < f(c-\epsilon_1)$ for decreasing functions. But since $g(x)=f(x)$ at both values, we also have $g(c-\epsilon_1) < u < g(c+\epsilon_1)$ for increasing functions, and $g(c+\epsilon_1) < u < g(c-\epsilon_1)$ for decreasing functions. | Then we add $L$ to each expression, and replace $f(x)$ by $g(x)$ everywhere. |

Both of these inequalities will lead to $c-\epsilon_1 < g^{-1}(u) < c+\epsilon_1$, because the direction of the inequalities will switch when applying a decreasing function to each expression of the inequality. | Then we apply the function $g^{-1}(x)$ to each expression. When $f(x)$ is a strictly increasing function, then both $g(x)$ and its inverse will also be strictly increasing, and the inequalities will not reverse direction. But when $f(x)$ is a strictly decreasing function, then both $g(x)$ and its inverse will also be strictly decreasing, and the direction of the inequality signs will change. |

Then $-\epsilon_1 < g^{-1}(u)-c < \epsilon_1$, which is equivalent to $|g^{-1}(u)-c| <\epsilon_1$, which implies $|g^{-1}(u)-c| <\epsilon$. | We then subtract $c$ from each term, rewrite as an absolute value inequality, then apply the definition of $\epsilon_1$ to recover $\epsilon$. |

Therefore, $\lim\limits_{u\to L} g^{-1}(u)=c$. | And our proof is complete. |