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near

it
may be to B at any instant, the next second it will

pass over one-half the interval still remaining; it must, therefore, approach nearer to B, since half the interval still remaining is some distance, but will not reach B, since half the interval still remaining is not the whole distance.

Hence, the distance from A to the moving point is an increasing variable, which indefinitely approaches the constant A B as its limit; and the distance from the moving point to B is a decreasing variable, which indefinitely approaches the constant zero as its limit.

If the length of A B be two inches, and the variable be denoted by x, and the difference between the variable and its limit, by v: after one second, x=1,

v=1; after two seconds, x=1+1,

v=}; after three seconds, X 1+1+1, = 1; after four seconds, 1+1+1+], v = $;

and so on indefinitely.

Now the sum of the series 1+1+1+ $ etc., is evidently less than 2; but by taking a great number of terms, the sum can be made to differ from 2 by as little as we please. Hence 2 is the limit of the sum of the series, when the number of the terms is increased indefinitely; and 0 is the limit of the variable difference between this variable sum and 2.

lim. will be used as an abbreviation for limit.

199. [1] The difference between a variable and its limit is a variable whose limit is zero.

[2] If two or more variables, v, v', v", etc., have zero for a limit, their sum, v tv! +0", etc., will have zero for a limit.

[3] If the limit of a variable, v, be zero, the limit of a + v will be the constant a, and the limit of a xv will be zero.

[4] The product of a constant and a variable is also a variable, and the limit of the product of a constant and a variable is the product of the constant and the limit of the variable.

[5] The sum or product of two variables, both of which are either increasing or decreasing, is also a variable.

PROPOSITION I.

C't

PL

[6] If two variables be always equal, their limits are equal.

Let the two variables A M and AN be always equal, and let AC

A and A B be their respective limits.

M M

N
We are to prove

AC= A B.
Suppose AC > A B. Then we may
diminish AC to some value AC such
that A C= A B.
Since A M approaches indefinitely to c

B A C, we may suppose that it has reached a value A P greater than A C'.

Let A Q be the corresponding value of A N.
Then

AP= A Q.
Now

AC = A B. But both of these equations cannot be true, for A P > A C', and A Q<A B... A C cannot be greater than A B.

Again, suppose AC< A B. Then we may diminish A B to some value A B' such that AC= A B'.

Since A N approaches indefinitely to A B we may suppose that it has reached a value A Q greater than A B'.

Let A P be the corresponding value of A M.
Then

A P= AQ.
Now

AC= A B'. But both of these equations cannot be true, for A P< A C, and AQ> A B'. .. A C cannot be less than A B.

Since A C cannot be greater or less than A B, it must be equal to A B.

[7] COROLLARY 1. If two variables be in a constant ratio, their limits are in the same ratio. For, let x and y be two variables having the constant ratio r, then

1", or, x =ry, therefore Y

lim. (x) lim. (x) = lim. (r y)=r X lim. (y), therefore

Q.E.D.

lim. (y)

[8] Cor. 2. Since an incommensurable ratio is the limit of its successive approximate values, two incommensurable ratios

6 a' and are equal if they always have the same approximate values

61 when expressed within the same measure of precision.

PROPOSITION II.

a

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[9] The limit of the algebraic sum of two or more variables is the algebraic sum of their limits.

Let X, Y, Z, be variables, a, b, and c, their respective limits, and v, w, and v'', the variable differences between x, y, z, b and a, b, c, respectively.

We are to prove lim. (x+y+z)=a+b+c.

Now, x=a-v, y=b- v, z=c-V. Then, x + y +z=a-v+b +c-v.

.. lim.(x+y+z)=lim.(a-v+b+c). [6] But, lim. (a v +b - v tc-") =a +b+c. [3]

lim. (x + y + 2) = a + b + c.

с

Q.E.D.

PROPOSITION III. [10] The limit of the product of two or more variables is the product of their limits.

Let X, Y, Z, be variables, a, b, c, their respective limits, and v, v, v', the variable differences between X, Y, Z, and a, b, c, respectively.

We are to prove lim. (x y z) = a b c.
Now, x = a - v, y=b V', z=(-0.
Multiply these equations together.

Then, x y z=abc F terms which contain one or more of the factors v, v, v'', and hence have zero for a limit.

[3] .. lim. (s y z) =lim. (abc F terms whose limits are zero). [6] But lim. (a b c 7 terms whose limits are zero) = a b c. .. lim. (x y z) = abc.

Q. E, D. For decreasing variables the proofs are similar.

Note. - In the application of the principles of limits, reference to this section (ş 199) will always include the fundamental truth of limits contained in Proposition I. ; and it will be left as an exercise for the student to determine in each case what other truths of this section, if any, are included in the reference.

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200. In the same circle, or equal circles, two commensurable arcs have the same ratio as the angles which they subtend at the centre.

B f

H K

e

P In the circle APC let the two arcs be A B and AC,

and AOB and AOC the which they subtend.

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Let H K be a common measure of A B and A C.
Suppose H K to be contained in A B three times,

and in AC five times.

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At the several points of division on A B and A C draw radii.

These radii will divide Z AOC into five equal parts, of which ZA O B will contain three,

$ 180 (in the same O, or equal TM, equal arcs subtend equal & at the centre).

LAOB 3
2 АОС 5

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arc A B arc AC

2 А ОВ
LA00'

Ax. 1.

Q. E. D.

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201. In the same circle, or in equal circles, incommensurable arcs have the same ratio as the angles which they subtend at the centre. P!

Р

CI

A

BI

B

D

arc AB

In the two equal © ABP and A'B'P' let A B and A' B'

be two incommensurable arcs, and C, C the is which they subtend at the centre.

arc A' B'

LC
We are to prove

Z C Let A B be divided into any number of equal parts, and let one of these parts be applied to A' B' as often as it will be contained in A'B'.

Since A B and A' B' are incommensurable, a certain number of these parts will extend from A to some point, as D, leaving a remainder D B' less than one of these parts.

Draw C'D.
Since A B and A'D are commensurable,
LA'C'D

$ 200

LACB' (two commensurable arcs have the same ratio as the which they subtend at

the centre). Now

suppose the number of parts into which A B is divided to be continually increased; then the length of each part will become less and less, and the point D will approach nearer and nearer to B', that is, the arc A' D will approach the arc A' B' as its limit, and the LA C'D the 2 A'C' B' as its limit.

arc A'D arc A B

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