# Elements of the Differential and Integral Calculus/Chapter III

CHAPTER III
THEORY OF LIMITS
$\scriptstyle {v}$
$\scriptstyle {l}$
$\scriptstyle {|v-l|}$
$\scriptstyle {v}$
$\scriptstyle {\mathbf {limit~} v=l,\mathbf {~or,~} v\doteq l}$.
The following familiar examples illustrate what is meant:
(1) As the number of sides of a regular inscribed polygon is indefinitely increased, the limit of the area of the polygon is the area of the circle. In this case the variable is always less than its limit.
(2) Similarly, the limit of the area of the circumscribed polygon is also the area of the circle, but now the variable is always greater than its limit.
(3) Consider the series
$\scriptstyle {1-{\frac {1}{2}}+{\frac {1}{4}}-{\frac {1}{8}}+\cdots }$
The sum of any even number $\scriptstyle {(2n)}$ of the first terms of this series is
| (B) | ${\begin{aligned}&\scriptstyle {S_{2n}=1-{\frac {1}{2}}+{\frac {1}{4}}-{\frac {1}{8}}+\cdots +{\frac {1}{2^{2n-2}}}-{\frac {1}{2^{2n-1}}},}\\&\scriptstyle {S_{2n}={\frac {{\frac {1}{2^{2n}}}-1}{-{\frac {1}{2}}-1}}={\frac {2}{3}}-{\frac {1}{3\cdot 2^{2n-1}}}.}\end{aligned}}$ | By 6, p. 1 |
Similarly, the sum of any odd number $\scriptstyle {(2n+1)}$ of the first terms of the series is
| (C) | ${\begin{aligned}&\scriptstyle {S_{2n+1}=1-{\frac {1}{2}}+{\frac {1}{4}}-{\frac {1}{8}}+\cdots -{\frac {1}{2^{2n-1}}}+{\frac {1}{2^{2n}}},}\\&\scriptstyle {S_{2n+1}={\frac {-{\frac {1}{2^{2n+1}}}-1}{-{\frac {1}{2}}-1}}={\frac {2}{3}}+{\frac {1}{3\cdot 2^{2n}}}.}\end{aligned}}$ | By 6, p. 1 |
$\scriptstyle {{\frac {2}{3}}-S_{2n}={\frac {1}{3\cdot 2^{2n-1}}},\quad S_{2n+1}-{\frac {2}{3}}={\frac {1}{3\cdot 2^{2n}}}}$,
$\scriptstyle {{\underset {n=\infty }{\operatorname {limit} }}\left({\frac {2}{3}}-S_{2n}\right)={\underset {n=\infty }{\operatorname {limit} }}{\frac {1}{3\cdot 2^{2n-1}}}=0}$
$\scriptstyle {{\underset {n=\infty }{\operatorname {limit} }}\left(S_{2n+1}-{\frac {2}{3}}\right)={\underset {n=\infty }{\operatorname {limit} }}{\frac {1}{3\cdot 2^{2n}}}=0}$
Hence, by definition of the limit of a variable, it is seen that both $\scriptstyle {S_{2n}}$ and $\scriptstyle {S_{2n+1}}$ are variables approaching $\scriptstyle {\frac {2}{3}}$ as a limit as the number of terms increases without limit.
Summing up the first two, three, four, etc., terms of (A), the sums are found by (B) and (C) to be alternately less and greater than $\scriptstyle {\frac {2}{3}}$, illustrating the case when the variable, in this case the sum of the terms of (A), is alternately less and greater than its limit.
In the examples shown the variable never reaches its limit. This is not by any means always the case, for from the definition of the limit of a variable it is clear that the essence of the definition is simply that the numerical value of the difference between the variable and its limit shall ultimately become and remain less than any positive number we may choose, however small.
$\scriptstyle {p_{n}}$
$\scriptstyle {P_{n}}$
$\scriptstyle {n}$
$\scriptstyle {C}$
$\scriptstyle {x}$
$\scriptstyle {P_{n},\quad p_{n+1},\quad C,\quad P_{n+1},\quad p_{n+2},\quad C,\quad P_{n+2},\quad }$ etc.
$\scriptstyle {{\underset {n=\infty }{\operatorname {limit} }}\;x=C}$
and the limit is reached by the variable, every third value of the variable being C.
14. Division by zero excluded. $\scriptstyle {\frac {0}{0}}$ is indeterminate. For the quotient of two numbers is that number which multiplied by the divisor will give the dividend. But any number whatever multiplied by zero gives ​zero, and the quotient is indeterminate; that is, any number whatever may be considered as the quotient, a result which is of no value.
$\scriptstyle {\frac {a}{0}}$ has no meaning, $\scriptstyle {a}$ being different from zero, for there exists no number such that if it be multiplied by zero, the product will equal $\scriptstyle {a}$.
Therefore division by zero is not an admissible operation.
Care should be taken not to divide by zero inadvertently. The following fallacy is an illustration.
${\begin{array}{lcrcl}\scriptscriptstyle {\quad {\text{Assume that}}}&\qquad &\scriptscriptstyle {a}&\scriptscriptstyle {=}&\scriptscriptstyle {b.}\\\scriptscriptstyle {\quad {\text{Then evidently}}}&&\scriptscriptstyle {ab}&\scriptscriptstyle {=}&\scriptscriptstyle {a^{2}.}\\\scriptscriptstyle {\quad {\text{Subtracting }}b^{2},}&&\scriptscriptstyle {ab-b^{2}}&\scriptscriptstyle {=}&\scriptscriptstyle {a^{2}-b^{2}.}\\\scriptscriptstyle {\quad {\text{Factoring,}}}&&\scriptscriptstyle {b(a-b)}&\scriptscriptstyle {=}&\scriptscriptstyle {(a+b)(a-b).}\\\scriptscriptstyle {\quad {\text{Dividing by }}a-b,}&&\scriptscriptstyle {b}&\scriptscriptstyle {=}&\scriptscriptstyle {a+b.}\\\scriptscriptstyle {\quad {\text{But}}}&&\scriptscriptstyle {a}&\scriptscriptstyle {=}&\scriptscriptstyle {b,}\\\scriptscriptstyle {\text{therefore}}&&\scriptscriptstyle {b}&\scriptscriptstyle {=}&\scriptscriptstyle {2b,}\\\scriptscriptstyle {\text{or,}}&&\scriptscriptstyle {1}&\scriptscriptstyle {=}&\scriptscriptstyle {2.}\end{array}}$
The result is absurd, and is caused by the fact that we divided by $\scriptscriptstyle {a-b=0}$.
$\scriptstyle {v}$
$\scriptstyle {\operatorname {limit} \;v=0}$, or, $\scriptstyle {v\doteq 0}$,
$\scriptstyle {v}$
$\scriptstyle {\operatorname {limit} \;v=l}$
$\scriptstyle {\operatorname {limit} \;(v-l)=0}$
Conversely, if the difference between a variable and a constant is an infinitesimal, then the variable approaches the constant as a limit.
$\scriptstyle {v}$
$\scriptstyle {\operatorname {limit\;} v=+\infty }$, or, $\scriptstyle {v\doteq +\infty }$.
$\scriptstyle {v}$
$\scriptstyle {\operatorname {limit\;} v=-\infty }$, or, $\scriptstyle {v\doteq -\infty }$.
$\scriptstyle {v}$
$\scriptstyle {\operatorname {limit\;} v=\infty }$, or, $\scriptstyle {v\doteq \infty }$.
Infinity ($\scriptstyle {\infty }$) is not a number; it simply serves to characterize a particular mode of variation of a variable by virtue of which it increases or decreases without limit.
17. Limiting value of a function. Given a function $\scriptstyle {f(x)}$.
$\scriptstyle {x}$
$\scriptstyle {\operatorname {limit\;} x=a}$,
$\scriptstyle {f(x)}$
$\scriptstyle {\operatorname {limit\;} f(x)=A}$,
$\scriptstyle {{\underset {x=a}{\operatorname {limit} }}\;f(x)=A}$,
and is read the limit of $\scriptstyle {f(x)}$, as $\scriptstyle {x}$ approaches the limit $\scriptstyle {a}$ in any manner, is $\scriptstyle {A}$.
$\scriptstyle {f(x)}$
$\scriptstyle {x}$
$\scriptstyle {a}$
$\scriptstyle {x=a}$
$\scriptstyle {{\underset {x=a}{\operatorname {limit} }}f(x)=f(a)}$,
$\scriptstyle {f(x)}$
$\scriptstyle {x=a}$
$\scriptstyle {x=a}$
$\scriptstyle {{\underset {x=a}{\operatorname {limit} }}f(x)=\infty }$,
the function is discontinuous for $\scriptstyle {x=a}$.
The attention of the student is now called to the following cases which occur frequently.
$\scriptstyle {f(x)={\frac {x^{2}-4}{x-2}}}$.
For $\scriptstyle {x=1}$, $\scriptstyle {f(x)=f(1)=3}$. Moreover, if $\scriptstyle {x}$ approaches the limit $\scriptstyle {1}$ in any manner, the function $\scriptstyle {f(x)}$ approaches $\scriptstyle {3}$ as a limit. Hence the function is continuous for $\scriptstyle {x=1}$.
Case II. The definition of a continuous function assumes that the function is already defined for $\scriptstyle {x=a}$. If this is not the case, however, it is sometimes possible to assign such a value to the function for $\scriptstyle {x=a}$ that the condition of continuity shall be satisfied. The following theorem covers these cases.
$\scriptstyle {{\underset {x=a}{\operatorname {limit} }}\;f(x)=B}$,
$\scriptstyle {\frac {x^{2}-4}{x-2}}$
$\scriptstyle {x=2}$
$\scriptstyle {x}$
$\scriptstyle {{\frac {x^{2}-4}{x-2}}=x+2}$;
$\scriptstyle {{\underset {x=2}{\operatorname {limit} }}\;(x+2)=4}$
$\scriptstyle {{\underset {x=2}{\operatorname {limit} }}\;{\frac {x^{2}-4}{x-2}}=4}$
Although the function is not defined for $\scriptstyle {x=2}$, if we arbitrarily assign it the value $\scriptstyle {4}$ for $\scriptstyle {x=2}$, it then becomes continuous for this value.
A function $\scriptstyle {f(x)}$ is said to be continuous in an interval when it is continuous for all values of $\scriptstyle {x}$ in this interval.
​19. Continuity and discontinuity of functions illustrated by their graphs.
(1) Consider the function $\scriptstyle {x^{2}}$, and let :
| (A) | $\scriptstyle {y=x^{2}}$. |
If we assume values for $\scriptstyle {x}$ and calculate the corresponding values of $\scriptstyle {y}$, we can plot a series of points. Drawing a smooth line free-hand through these points, a good representation of the general behavior of the function may be obtained. This picture or image of the function is called its graph. It is evidently the locus of all points satisfying equation (A).
Such a series or assemblage of points is also called a curve. Evidently we may assume values of $\scriptstyle {x}$ so near together as to bring the values of $\scriptstyle {y}$ (and therefore the points of the curve) as near together as we please. In other words, there are no breaks in the curve, and the function $\scriptstyle {x^{2}}$ is continuous for all values of $\scriptstyle {x}$.
$\scriptstyle {\sin x}$
$\scriptstyle {y=\sin x}$.
It is seen that no break in the curve occurs anywhere.
(3) The continuous function $\scriptstyle {e^{x}}$ is of very frequent occurrence in the Calculus. If we plot its graph from$\scriptstyle {(e=2.718\cdots )}$$\scriptstyle {y=e^{x},}$we get a smooth curve as shown. From this it is clearly seen that,
(a) when $\scriptstyle {x=0}$, $\scriptstyle {{\underset {x=0}{\operatorname {limit} }}\;y(=e^{x})=1}$;
(b) when $\scriptstyle {x>0}$, $\scriptstyle {y(=e^{x})}$ is positive and increases as we pass towards the right from the origin;
(c) when $\scriptstyle {x<0}$, $\scriptstyle {y(=e^{x})}$ is still positive and decreases as we pass towards the left from the origin.
(4) The function $\scriptstyle {\log _{e}x}$ is closely related to the last one discussed. In fact, if we plot its graph from$\scriptstyle {y=\log _{e}x,}$it will be seen that its graph has the same relation to $\scriptstyle {OX}$ and $\scriptstyle {OY}$ as the graph of $\scriptstyle {e^{x}}$ has to $\scriptstyle {OY}$ and $\scriptstyle {OX}$.
​Here we see the following facts pictured:
(a) For $x=1$, $\log _{e}x=log_{e}1=0$.
(b) For $x>1$, $\log _{e}x$ is positive and increases as $x$ increases.
(c) For $1>x>0$, $\log _{e}x$ is negative and increases in numerical value as $x$ diminishes, that is, ${\underset {x=0}{\operatorname {limit} }}\;\log x=-\infty$.
(d) For $x\eqslantless 0$, $log_{e}x$ is not defined; hence the entire graph lies to the right of $OY$.
(5) Consider the function ${\frac {1}{x}}$, and set $y={\frac {1}{x}}.$
If the graph of this function be plotted, it will be seen that as $x$ approaches the value zero from the left (negatively), the points of the curve ultimately drop down an infinitely great distance, and as $x$ approaches the value zero from the right, the curve extends upward infinitely far.
The curve then does not form a continuous branch from one side to the other of the axis of $Y$, showing graphically that the function is discontinuous for $x=0$, but continuous for all other values of $x$.
(6) From the graph of $y={\frac {2x}{1-x^{2}}}$it is seen that the function ${\frac {2x}{1-x^{2}}}$is discontinuous for the two values $x=\pm 1$, but continuous for all other values of $x$.
(7) The graph of $y=\tan x$shows that the function $\tan x$ is discontinuous for infinitely many values of the independent variable $x$, namely, $x={\frac {n\pi }{2}}$, where $n$ denotes any odd positive or negative integer.
(8) The function $\operatorname {arc~tan} x$ ​has infinitely many values for a given value of $\scriptstyle {x}$, the graph of equation$\scriptstyle {y=\operatorname {arc~tan} x}$consisting of infinitely many branches. If, however, we confine ourselves to any single branch, the function is continuous. For instance, if we say that $\scriptstyle {y}$ shall be the arc of smallest numerical value whose tangent is $\scriptstyle {x}$, that is, $\scriptstyle {y}$ shall take on only values between $\scriptstyle {-{\frac {\pi }{2}}}$ and $\scriptstyle {\frac {\pi }{2}}$, then we are limited to the branch passing through the origin, and the condition for continuity is satisfied.
(9) Similarly, $\scriptstyle {\operatorname {arc~tan} {\frac {1}{x}},}$is found to be a many-valued function. Confining ourselves to one branch of the graph of $\scriptstyle {y=\operatorname {arc~tan} {\frac {1}{x}},}$we see that as $\scriptstyle {x}$ approaches zero from the left, $\scriptstyle {y}$ approaches the limit $\scriptstyle {-{\frac {\pi }{2}},}$ and as $\scriptstyle {x}$ approaches zero from the right, $\scriptstyle {y}$ approaches the limit $\scriptstyle {+{\frac {\pi }{2}}.}$ Hence the function is discontinuous when $\scriptstyle {x=0.}$ Its value for $\scriptstyle {x=0}$ can be assigned at pleasure.
Functions exist which are discontinuous for every value of the independent variable within a certain range. In the ordinary applications of the Calculus, however, we deal with functions which are discontinuous (if at all) only for certain isolated values of the independent variable; such functions are therefore in general continuous, and are the only ones considered in this book.
20. 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 I. The limit of the algebraic sum of a finite number of variables is equal to the like algebraic sum of the limits of the several variables.
Theorem II. The limit of the product of a finite number of variables is equal to the product of the limits of the several variables.
Theorem III. 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.
​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 value of this sum can be made less than any small positive quantity (as $\scriptstyle {\epsilon }$) that may be assigned (§15). That this is possible is evident, for, the limit of each infinitesimal being zero, each one can be made numerically less than $\scriptstyle {\frac {\epsilon }{n}}$ ($\scriptstyle {n}$ being the number of infinitesimals), and therefore their sum can be made numerically less than $\scriptstyle {\epsilon }$.
(2) The product of a constant $\scriptstyle {c}$ and an infinitesimal is an infinitesimal. For the numerical value of the product can always be made less than any small positive quantity (as $\scriptstyle {\epsilon }$) by making the numerical value of the infinitesimal less than $\scriptstyle {\frac {\epsilon }{c}}$.
(3) 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 $\scriptstyle {n}$ factors, then since each infinitesimal may be assumed less than the $\scriptstyle {n}$th root of $\scriptstyle {\epsilon }$, the product can be made less than $\scriptstyle {\epsilon }$ itself.
(4) If $\scriptstyle {v}$ is a variable which approaches a limit $\scriptstyle {l}$ different from zero, then the quotient of an infinitesimal by $\scriptstyle {v}$ is also an infinitesimal. For if limit $\scriptstyle {v=l}$, and $\scriptstyle {k}$ is any number numerically less than $\scriptstyle {l}$, then, by definition of a limit, $\scriptstyle {v}$ will ultimately become and remain numerically greater than $\scriptstyle {k}$. Hence the quotient $\scriptstyle {\frac {\epsilon }{v}}$, where $\scriptstyle {\epsilon }$ is an infinitesimal, will ultimately become and remain numerically less than $\scriptstyle {\frac {\epsilon }{k}}$, and is therefore by (2) an infinitesimal.
$\scriptstyle {v_{1},v_{2},v_{3},\cdots }$
$\scriptstyle {l_{1},l_{2},l_{3},\cdots }$
${\begin{array}{c}\scriptstyle {v_{1}-l_{1}=\epsilon _{1},}\\\scriptstyle {v_{2}-l_{2}=\epsilon _{2},}\\\scriptstyle {v_{3}-l_{3}=\epsilon _{3},}\\\scriptstyle {.~.~.~.~.}\end{array}}$
$\scriptstyle {\epsilon _{1},\epsilon _{2},\epsilon _{3},\cdots }$
$\scriptstyle {(v_{1}+v_{2}+v_{3}+\cdots )-(l_{1}+l_{2}+l_{3}+\cdots )=(\epsilon _{1}+\epsilon _{2}+\epsilon _{3}+\cdots ).}$
​Since the right-hand member is an infinitesimal by (1), p. 19, we have, from the converse theorem on p. 18,
| or, | ${\begin{aligned}&\scriptstyle {\operatorname {limit\;} (v_{1}+v_{2}+v_{3}+\cdots )=l_{1}+l_{2}+l_{3}+\cdots ,}\\&\scriptstyle {\operatorname {limit\;} (v_{1}+v_{2}+v_{3}+\cdots )=\operatorname {limit\;} v_{1}+\operatorname {limit\;} v_{2}+\operatorname {limit\;} v_{3}+\cdots ,}\end{aligned}}$ |
which was to be proved.
Proof of Theorem II. Let $\scriptstyle {v_{1}}$ and $\scriptstyle {v_{2}}$ be the variables, $\scriptstyle {l_{1}}$ and $\scriptstyle {l_{2}}$ their respective limits, and $\scriptstyle {\epsilon _{1}}$ and $\scriptstyle {\epsilon _{2}}$ infinitesimals; then
| and | ${\begin{aligned}&\scriptstyle {v_{1}=l_{1}+\epsilon _{1}}\\&\scriptstyle {v_{2}=l_{2}+\epsilon _{2}.}\end{aligned}}$ | and |
| Multiplying, | ${\begin{aligned}\scriptstyle {v_{1}v_{2}}&\scriptstyle {=(l_{1}+\epsilon _{1})(l_{2}+\epsilon _{2})}\\&\scriptstyle {=l_{1}l_{2}+l_{1}\epsilon _{2}+l_{2}\epsilon _{1}+\epsilon _{1}\epsilon _{2},}\end{aligned}}$ | Multiplying, |
$\scriptstyle {v_{1}v_{2}-l_{1}l_{2}=l_{1}\epsilon _{2}+l_{2}\epsilon _{1}+\epsilon _{1}\epsilon _{2}.}$
Since the right-hand member is an infinitesimal by (1) and (2), p. 19, we have, as before,$\scriptstyle {\operatorname {limit\;} (v_{1}v_{2})=l_{1}l_{2}=\operatorname {limit\;} v_{1}\cdot \operatorname {limit\;} v_{2},}$which was to be proved.
$\scriptstyle {{\frac {v_{1}}{v_{2}}}={\frac {l_{1}+\epsilon _{1}}{l_{2}+\epsilon _{2}}}={\frac {l_{1}}{l_{2}}}+\left({\frac {l_{1}+\epsilon _{1}}{l_{2}+\epsilon _{2}}}-{\frac {l_{1}}{l_{2}}}\right),}$
$\scriptstyle {{\frac {v_{1}}{v_{2}}}-{\frac {l_{1}}{l_{2}}}={\frac {l_{2}\epsilon _{1}-l_{1}\epsilon _{2}}{l_{2}(l_{2}+\epsilon _{2})}}.}$
Here again the right-hand member is an infinitesimal by (4), p. 19, if $\scriptstyle {l_{2}\neq 0}$; hence$\scriptstyle {\operatorname {limit\;} \left({\frac {v_{1}}{v_{2}}}\right)={\frac {l_{1}}{l_{2}}}={\frac {\operatorname {limit\;} v_{1}}{\operatorname {limit\;} v_{2}}},}$which was to be proved.
It is evident that if any of the variables be replaced by constants, our reasoning still holds, and the above theorems are true.
21. Special limiting values. The following examples are of special importance in the study of the Calculus. In the following examples $\scriptstyle {a>0}$ and $\scriptstyle {c\neq 0}$.
| Written in the form of limits. | Abbreviated form often used. |
| ${\begin{array}{rrclrcl}\scriptstyle {{\text{(1)}}~}&\scriptstyle {{\underset {x=0}{\operatorname {limit} }}\;{\frac {c}{x}}}&\scriptstyle {=}&\scriptstyle {\infty ;}&\scriptstyle {\frac {c}{0}}&\scriptstyle {=}&\scriptstyle {\infty .}\\\scriptstyle {{\text{(2)}}~}&\scriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;cx}&\scriptstyle {=}&\scriptstyle {\infty ;}&\scriptstyle {c\cdot \infty }&\scriptstyle {=}&\scriptstyle {\infty .}\\\scriptstyle {{\text{(3)}}~}&\scriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {x}{c}}}&\scriptstyle {=}&\scriptstyle {\infty ;}&\scriptstyle {\frac {\infty }{c}}&\scriptstyle {=}&\scriptstyle {\infty .}\\\scriptstyle {{\text{(4)}}~}&\scriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {c}{x}}}&\scriptstyle {=}&\scriptstyle {0;}&\scriptstyle {\frac {c}{\infty }}&\scriptstyle {=}&\scriptstyle {0.}\\\scriptstyle {{\text{(5)}}~}&\scriptstyle {{\underset {x=-\infty }{\operatorname {limit} }}\;a^{x}}&\scriptstyle {=}&\scriptstyle {+\infty ,{\text{ when }}a<1;}&\scriptstyle {a^{-\infty }}&\scriptstyle {=}&\scriptstyle {+\infty .}\\\scriptstyle {{\text{(6)}}~}&\scriptstyle {{\underset {x=+\infty }{\operatorname {limit} }}\;a^{x}}&\scriptstyle {=}&\scriptstyle {0,\quad {\text{ when }}a<1;}&\scriptstyle {a^{+\infty }}&\scriptstyle {=}&\scriptstyle {0.}\\\scriptstyle {{\text{(7)}}~}&\scriptstyle {{\underset {x=-\infty }{\operatorname {limit} }}\;x^{x}}&\scriptstyle {=}&\scriptstyle {0,\quad {\text{ when }}a>1;}&\scriptstyle {a^{-\infty }}&\scriptstyle {=}&\scriptstyle {0.}\\\scriptstyle {{\text{(8)}}~}&\scriptstyle {{\underset {x=+\infty }{\operatorname {limit} }}\;a^{x}}&\scriptstyle {=}&\scriptstyle {+\infty {\text{ when}}a>1;}&\scriptstyle {a^{+\infty }}&\scriptstyle {=}&\scriptstyle {+\infty .}\\\scriptstyle {{\text{(9)}}~}&\scriptstyle {{\underset {x=0}{\operatorname {limit} }}\;\log _{a}x}&\scriptstyle {=}&\scriptstyle {+\infty {\text{ when }}a<1;}&\scriptstyle {\log _{a}0}&\scriptstyle {=}&\scriptstyle {+\infty .}\\\scriptstyle {{\text{(10)}}~}&\scriptstyle {{\underset {x=+\infty }{\operatorname {limit} }}\;\log _{a}x}&\scriptstyle {=}&\scriptstyle {-\infty ,{\text{ when}}a<1;}&\scriptstyle {\log _{a}(+\infty )}&\scriptstyle {=}&\scriptstyle {-\infty .}\\\scriptstyle {{\text{(11)}}~}&\scriptstyle {{\underset {x=0}{\operatorname {limit} }}\;\log _{a}x}&\scriptstyle {=}&\scriptstyle {-\infty ,{\text{ when }}a>1;}&\scriptstyle {\log _{a}0}&\scriptstyle {=}&\scriptstyle {-\infty .}\\\scriptstyle {{\text{(12)}}~}&\scriptstyle {{\underset {x=+\infty }{\operatorname {limit} }}\;\log _{a}x}&\scriptstyle {=}&\scriptstyle {+\infty ,{\text{ when }}a>1;}&\scriptstyle {\log _{a}(+\infty )}&\scriptstyle {=}&\scriptstyle {+\infty .}\end{array}}$ |||
The expressions in the second column are not to be considered as expressing numerical equalities ($\scriptstyle {\infty }$ not being a number); they are merely symbolical equations implying the relations indicated in the first column, and should be so understood.
22. Show that $\scriptstyle {{\underset {x=0}{\operatorname {limit} }}\;{\frac {\sin x}{x}}=1}$.
Let $\scriptstyle {O}$ be the center of a circle whose radius is unity.
Let $\scriptstyle {\operatorname {arc~} AM=\operatorname {arc~} AM'=x}$, and let $\scriptstyle {MT}$ and $\scriptstyle {M'T}$ be tangents drawn to the circle at $\scriptstyle {M}$ and $\scriptstyle {M'}$. From Geometry,
| | $\scriptstyle {MPM'<MAM'<MTM'}$; |
| or | $\scriptstyle {2\sin x<2x<2\tan x}$. |
Dividing through by $\scriptstyle {2\sin x}$, we get$\scriptstyle {1<{\frac {x}{\sin x}}<{\frac {1}{\cos x}}.}$ ​If now $x$ approaches the limit zero,${\underset {x=0}{\operatorname {limit} }}\;{\frac {x}{\sin x}}$must lie between the constant $1$ and ${\underset {x=0}{\operatorname {limit} }}{\frac {1}{\cos x}}$, which is also $1$.
| Therefore ${\underset {x=0}{\operatorname {limit} }}\;{\frac {x}{\sin x}}=1$, or, ${\underset {x=0}{\operatorname {limit} }}\;{\frac {\sin x}{x}}=1$. | Th. III, p. 18 |
$y={\frac {\sin x}{x}}.$
Although the function is not defined for $x=0$, yet it is not discontinuous when $x=0$ if we define
| ${\frac {\sin 0}{0}}=1.$ | Case II, p. 15 |
23. The number $\mathbf {e}$. One of the most important limits in the Calculus is${\underset {x=0}{\operatorname {limit} }}\;(1+x)^{\frac {1}{x}}=2.71828\cdots =e.$
To prove rigorously that such a limit $e$ exists, is beyond the scope of this book. For the present we shall content ourselves by plotting the locus of the equation$y=(1+x)^{\frac {1}{x}}$and show graphically that, as $x\doteq 0$, the function $(1+x)^{\frac {1}{x}}(=y)$
| $x$ | $y$ | $x$ | $y$ |
| 10 | 1.0096 | | |
| 5 | 1.4310 | | |
| 2 | 1.7320 | | |
| 1 | 2.0000 | | |
| .5 | 2.2500 | —.5 | 4.0000 |
| .1 | 2.5937 | —.1 | 2.8680 |
| .01 | 2.7048 | —.01 | 2.7320 |
| .001 | 2.7169 | —.001 | 2.7195 |
takes on values in the near neighborhood of $2.718\cdots$, and therefore $e=2.718\cdots$ approximately.
​As $\scriptstyle {x\doteq 0}$ from the left, $\scriptstyle {y}$ decreases and approaches $\scriptstyle {e}$ as a limit. As $\scriptstyle {x\doteq 0}$ from the right, $\scriptstyle {y}$ increases and also approaches $\scriptstyle {e}$ as a limit.
As $\scriptstyle {x\doteq \infty }$, $\scriptstyle {y}$ approaches the limit $\scriptstyle {1}$; and as $\scriptstyle {\doteq -1}$ from the right, $\scriptstyle {y}$ increases without limit.
In Chap. XVIII, Ex. 15, p. 233, we will show how to calculate the value of $\scriptstyle {e}$ to any number of decimal places.
Natural logarithms are those which have the number $\scriptstyle {e}$ for base. These logarithms play a very important role in mathematics. When the base is not indicated explicitly, the base $\scriptstyle {e}$ is always understood in what follows in this book. Thus $\scriptstyle {\log _{e}v}$ is written simply $\scriptstyle {\log v}$.
Natural logarithms possess the following characteristic property: If $\scriptstyle {x\doteq 0}$ in any way whatever,$\scriptstyle {\operatorname {limit\;} {\frac {\log(1+x)}{x}}=\operatorname {limit\;} \log(1+x)^{\frac {1}{x}}=\log e=1.}$
24. Expressions assuming the form $\scriptstyle {\frac {\infty }{\infty }}$. As $\scriptstyle {\infty }$ is not a number, the expression $\scriptstyle {\infty \div \infty }$ is indeterminate. To evaluate a fraction assuming this form, the numerator and denominator being algebraic functions, we shall find useful the following
Rule. Divide both numerator and denominator by the highest power of the variable occurring in either. Then substitute the value of the variable.
Illustrative Example 1. Evaluate $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {2x^{3}-3x^{2}+4}{5x-x^{2}-7x^{3}}}}$.
Solution. Substituting directly, we get $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {2x^{3}-3x^{2}+4}{5x-x^{2}-7x^{3}}}={\frac {\infty }{\infty }}}$, which is indeterminate. Hence, following the above rule, we divide both numerator and denominator by $\scriptstyle {x^{3}}$. Then$\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\;{\frac {2x^{3}-3x^{2}+4}{5x-x^{2}-7x^{3}}}={\underset {x=\infty }{\operatorname {limit} }}\;{\frac {2-{\frac {3}{x}}+{\frac {4}{x^{3}}}}{{\frac {5}{x^{2}}}-{\frac {1}{x}}-7}}=-{\frac {2}{7}}.\quad Ans.}$
EXAMPLES
Prove the following:
1. $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {x+1}{x}}\right)=1}$.
| Proof. | ${\begin{aligned}\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {x+1}{x}}\right)}&\scriptscriptstyle {={\underset {x=\infty }{\operatorname {limit} }}\left(1+{\frac {1}{x}}\right)}\\&\scriptscriptstyle {={\underset {x=\infty }{\operatorname {limit} }}(1)+{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {1}{x}}\right)}\\&\scriptscriptstyle {=1+0=1.}\end{aligned}}$ | Th. I, p. 18 |
| Proof. | $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {x^{2}+2x}{5-3x^{2}}}\right)={\underset {x=\infty }{\operatorname {limit} }}\left({\frac {1+{\frac {2}{x}}}{{\frac {5}{x^{2}}}-3}}\right)}$ | |
| [Dividing both numerator and denominator by $\scriptscriptstyle {x^{2}}$.] |||
| Proof. | $\scriptscriptstyle {={\frac {{\underset {x=\infty }{\operatorname {limit} }}\left(1+{\frac {2}{x}}\right)}{{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {5}{x^{2}}}-3\right)}}}$ | Th. III, p. 18 |
| Proof. | $\scriptscriptstyle {={\frac {{\underset {x=\infty }{\operatorname {limit} }}(1)+{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {2}{x}}\right)}{{\underset {x=\infty }{\operatorname {limit} }}\left({\frac {5}{x^{2}}}\right)-{\underset {x=\infty }{\operatorname {limit} }}(3)}}}$ | Th. I, p. 18 |
| Proof. | $\scriptscriptstyle {={\frac {1+0}{0-3}}=-{\frac {1}{3}}.}$ | |
3. $\scriptscriptstyle {{\underset {x=1}{\operatorname {limit} }}{\frac {x^{2}-2x+5}{x^{2}+7}}={\frac {1}{2}}}$.
4. $\scriptscriptstyle {{\underset {x=0}{\operatorname {limit} }}{\frac {3x^{3}+6x^{2}}{2x^{4}-15x^{2}}}=-{\frac {2}{5}}}$.
5. $\scriptscriptstyle {{\underset {x=-2}{\operatorname {limit} }}{\frac {x^{2}+1}{x+3}}=5}$.
6. $\scriptscriptstyle {{\underset {h=0}{\operatorname {limit} }}(3ax^{2}-2hx+5h^{2})=3ax^{2}}$.
7. $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}(ax^{2}+bx+c)=\infty }$.
8. $\scriptscriptstyle {{\underset {k=0}{\operatorname {limit} }}{\frac {(x-k)^{2}-2kx^{3}}{x(x+k)}}=1}$.
9. $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}{\frac {x^{2}+1}{3x^{2}+2x-1}}={\frac {1}{3}}}$.
10. $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}{\frac {3+2x}{x^{2}-5x}}=0}$.
11. $\scriptscriptstyle {{\underset {\alpha ={\frac {\pi }{2}}}{\operatorname {limit} }}{\frac {\cos(\alpha -a)}{\cos(2\alpha -a)}}=-\tan a}$.
12. $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}{\frac {ax^{2}+bx+c}{dx^{2}+ex+f}}={\frac {a}{d}}}$.
13. $\scriptscriptstyle {{\underset {z=0}{\operatorname {limit} }}{\frac {a}{2}}(e^{\frac {z}{a}}+e^{-{\frac {z}{a}}})=a}$.
14. $\scriptscriptstyle {{\underset {x=0}{\operatorname {limit} }}{\frac {2x^{3}+3x^{2}}{x^{3}}}=\infty }$.
15. $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}{\frac {5x^{2}-2x}{x}}=\infty }$.
16. $\scriptscriptstyle {{\underset {y=\infty }{\operatorname {limit} }}{\frac {y}{y+1}}=1}$.
17. $\scriptscriptstyle {{\underset {n=\infty }{\operatorname {limit} }}{\frac {n(n+1)}{(n+2)(n+3)}}=1}$.
18. $\scriptscriptstyle {{\underset {s=1}{\operatorname {limit} }}{\frac {s^{3}-1}{s-1}}=3}$.
19. $\scriptscriptstyle {{\underset {h=0}{\operatorname {limit} }}{\frac {(x+h)^{n}-x^{n}}{h}}=nx^{n-1}}$.
20. $\scriptscriptstyle {{\underset {h=0}{\operatorname {limit} }}\left[\cos(\theta +h){\frac {\sin h}{h}}\right]=\cos \theta }$.
21. $\scriptscriptstyle {{\underset {x=\infty }{\operatorname {limit} }}{\frac {4x^{2}-x}{4-3x^{2}}}=-{\frac {4}{3}}}$.
22. $\scriptscriptstyle {{\underset {\theta =0}{\operatorname {limit} }}{\frac {1-\cos \theta }{\theta ^{2}}}={\frac {1}{2}}}$.
23. $\scriptscriptstyle {{\underset {x=a}{\operatorname {limit} }}{\frac {1}{x-a}}=-\infty }$, if $\scriptscriptstyle {x}$ is increasing as it approaches the value $\scriptscriptstyle {a}$.
24. $\scriptscriptstyle {{\underset {x=a}{\operatorname {limit} }}{\frac {1}{x-a}}=+\infty }$, if $\scriptscriptstyle {x}$ is decreasing as it approaches the value $\scriptscriptstyle {a}$.
* ↑ To be read the numerical value of the difference between $\scriptstyle {v}$ and $\scriptstyle {l}$.
* ↑ Hence a constant, no matter how small it may be, is not an infinitesimal.
* ↑ On account of the notation used and for the sake of uniformity, the expression $\scriptstyle {v\doteq +\infty }$ is sometimes read $\scriptstyle {v}$ approaches the limit plus infinity. Similarly, $\scriptstyle {v\doteq -\infty }$ is read $\scriptstyle {v}$ approaches the limit minus infinity, and $\scriptstyle {v\doteq \infty }$ is read $\scriptstyle {v}$, in numerical value, approaches the limit infinity.While the above notation is convenient to use in this connection, the student must not forget that infinity is not a limit in the sense in which we defined a limit on p. 11, for infinity is not a number at all.
* ↑ In this book we shall deal only with functions which are in general continuous, that is, continuous for all values of $\scriptstyle {x}$, with the possible exception of certain isolated values, our results in general being understood as valid only for such values of $\scriptstyle {x}$ for which the function in question is actually continuous. Unless special attention is called thereto, we shall as a rule pay no attention to the possibilities of such exceptional values of $\scriptstyle {x}$ for which the function is discontinuous. The definition of a continuous function $\scriptstyle {f(x)}$ is sometimes roughly (but imperfectly) summed up in the statement that a small change in $\scriptstyle {x}$ shall produce a small change in $\scriptstyle {f(x)}$. We shall not consider functions having an infinite number of oscillations in a limited region.
* ↑ If we refer to the table on p. 4, it will be seen that for all angles less than 10° the angle in radians and the sine of the angle are equal to three decimal places. If larger tables are consulted, five-place, say, it will be seen that for all three angles less than 2.2° the sine of the angle and the angle itself are equal to four decimal places. From this we may well suspect that $\scriptscriptstyle {{\underset {x=0}{\operatorname {limit} }}\;{\frac {\sin x}{x}}=1.}$