Fundamental theorems on limits

In problems involving limits the use of one or more of the following theorems is usually implied. It is assumed that the limit of each variable exists and is finite.

Theorem 3.8.1   The limit of the algebraic sum of a finite number of variables is equal to the algebraic sum of the limits of the several variables.

In particular,

$$\lim_{x\to a} [f(x)+g(x)] = \lim_{x\to a} f(x) +\lim_{x\to a} g(x).$$

Theorem 3.8.2   The limit of the product of a finite number of variables is equal to the product of the limits of the several variables.

In particular,

$$\lim_{x\to a} [f(x)\cdot g(x)] = \lim_{x\to a} f(x) \cdot \lim_{x\to a} g(x).$$

Theorem 3.8.3   The limit of the quotient of two variables is equal to the quotient of the limits of the separate variables, provided the limit of the denominator is not zero.

In particular,

$$\lim_{x\to a} [f(x)/g(x)] = \frac{\lim_{x\to a} f(x)}{\lim_{x\to a} g(x)},$$

provided $\lim_{x\to a} g(x)\not= 0$.

Before proving these theorems it is necessary to establish the following properties of infinitesimals.

1. The sum of a finite number of infinitesimals is an infinitesimal. To prove this we must show that the numerical^3.4value of this sum can be made less than any small positive quantity (as $\epsilon$) that may be assigned (§3.3). That this is possible is evident, for, the limit of each infinitesimal being zero, each one can be made numerically less than $\frac{\epsilon}{n}$ ($n$ being the number of infinitesimals), and therefore their sum can be made numerically less than $\epsilon$.

2. The product of a constant $c\not= 0$ and an infinitesimal is an infinitesimal. For the numerical value of the product can always be made less than any small positive quantity (as $\epsilon$) by making the numerical value of the infinitesimal less than $\frac{\epsilon}{\vert c\vert}$.

3. If $v$ is a variable which approaches a limit $L$ different from zero, then the quotient of an infinitesimal by $v$ is also an infinitesimal. For if $v\to L$, and $k$ is any number numerically less than $L$, then, by definition of a limit, $v$ will ultimately become and remain numerically greater than $k$. Hence the quotient $\frac{\epsilon}{v}$, where $\epsilon$ is an infinitesimal, will ultimately become and remain numerically less than $\frac{\epsilon}{k}$, and is therefore by the previous item an infinitesimal.

4. The product of any finite number of infinitesimals is an infinitesimal. For the numerical value of the product may be made less than any small positive quantity that can be assigned. If the given product contains $n$ factors, then since each infinitesimal may be assumed less than the $n-th$ root of $\epsilon$, the product can be made less than $\epsilon$ itself.

Proof of Theorem 3.8.1. Let $v_1$, $v_2$, $v_3$, $\dots$ be the variables, and $L_1$, $L_2$, $L_3$, $\dots$ their respective limits. We may then write

$$v_1 - L_1 = \epsilon_1,\ \ v_2 - L_2 = \epsilon_2,\ \ v_3 - L_3 = \epsilon_3,$$

where $\epsilon_1$, $\epsilon_2$, $\epsilon_3$, $\dots$ are infinitesimals (i.e. variables having zero for a limit). Adding

$$(v_1 + v _2 + v_3 + \dots) - (L_1 + L_2 + L_3 + . . .) = (\epsilon_1 + \epsilon_2 + \epsilon_3 + \dots).$$

Since the right-hand member is an infinitesimal by item (1) above (§3.8), we have, from the converse theorem (§3.3),

$$\lim (v_1 + v_2 + v_3 + \dots) = L_1 + L_2 + L_3 + \dots,$$

or,

$$\lim (v_1 + v_2 + v_3 + \dots) = \lim v_1 + \lim v_2 + \lim v_3 + \dots,$$

which was to be proved. $\qedsymbol$

Proof of Theorem 3.8.2. Let $v_1$ and $v_2$ be the variables, $L_1$ and $L_2$ their respective limits, and $\epsilon_1$ and $\epsilon_2$ infinitesimals; then

$$v_1 = L_1 + \epsilon_1$$

and $v_2 = L_2 + \epsilon_2$. Multiplying,

$$\begin{array} {ll} v_1v_2 &= (L_1 + \epsilon_1)(L_2 + \epsil... ..._1\epsilon_2 + L_2\epsilon_1 + \epsilon_1\epsilon_2 \end{array}$$

or,

$$v_1v_2 - L_1L_2 = L_1\epsilon_2 + L_2\epsilon_1 + \epsilon_1\epsilon_2.$$

Since the right-hand member is an infinitesimal by items (1) and (2) above, (§3.8), we have, as before,

$$\lim (v_1v_2) = L_1L_2 = \lim v_1 \cdot \lim v_2,$$

which was to be proved. $\qedsymbol$

Proof of Theorem 3.8.3. Using the same notation as before,

$$\frac{v_1}{v_2} = \frac{L_1 + \epsilon_1}{L_2 + \epsilon_2} = \f... ...\left ( \frac{L_1 + \epsilon_1}{L_2 + \epsilon_2} - \frac{L_1}{L_2} \right ),$$

or,

$$\frac{v_1}{v_2} - \frac{L_1}{L_2} = \frac{L_2 \epsilon_1 - L_1 \epsilon_2}{L_2 (L_2 + \epsilon_2)}.$$

Here again the right-hand member is an infinitesimal by item (3) above, (§3.8), if $L_2 \ne 0$; hence

$$\lim \left ( \frac{v_1}{v_2} \right ) = \frac{L_1}{L_2} = \frac{\lim\,v_1}{\lim\,v_2},$$

which was to be proved. $\qedsymbol$

It is evident that if any of the variables be replaced by constants, our reasoning still holds, and the above theorems are true.