Limiting Ratios of Magnitudes that Increase Without Limit

Excerpts

• Limiting Ratios of Magnitudes that Increase Without Limit

  It has been shown that by taking $\dfrac{1}{x}$ sufficiently small, that is, by taking $x$ sufficiently great, any term of this series may be made to contain the aggregate of the succeeding terms, as often as we please; which relation is not altered if we multiply every term by $x^{m}$, and so restore the original series.

• Limiting Ratios of Magnitudes that Increase Without Limit

  Hence $\dfrac{(x + 1)^{m}}{x^{m}} = 1 + \dfrac{mx^{m-1} + \etc.}{x^{m}}$, the numerator of which last fraction decreases indefinitely as compared with its denominator.

• Limiting Ratios of Magnitudes that Increase Without Limit

  The limit of any ratio may be found by rejecting any terms or aggregate of terms ($Q$) which are connected with another term ($P$) by the sign of addition or subtraction, provided that by increasing $x$, $Q$ may be made as small a part of $P$ as we please.

• Limiting Ratios of Magnitudes that Increase Without Limit

  Divide both numerator and denominator by $x^{2}$, which gives $1 + \dfrac{2}{x} + \dfrac{3}{x^{2}}$, and $2 + \dfrac{5}{x}$, for the numerator and denominator of a fraction equal in value to the one proposed.

• Limiting Ratios of Magnitudes that Increase Without Limit

  It is easy to show that the increase of two magnitudes may cause a decrease of their ratio; so that, as the two increase without limit, their ratio may diminish without limit.

• Limiting Ratios of Magnitudes that Increase Without Limit

  We will now prove the following: That in any series of decreasing powers of $x$, any one term will, if $x$ be taken sufficiently great, contain the aggregate of all which follow, as many times as we please.

• Limiting Ratios of Magnitudes that Increase Without Limit

  This result will be of use when we come to the first principles of the integral calculus.

Equations

• Limiting Ratios of Magnitudes that Increase Without Limit

  \dfrac{(x + 1)^{m}}{x^{m}} = 1 + \dfrac{mx^{m-1} + \etc.}{x^{m}}

  The ratio (x+1)^m / x^m equals 1 plus a remainder whose numerator becomes negligible next to x^m as x grows without limit, so the ratio tends to unity.

• Limiting Ratios of Magnitudes that Increase Without Limit

  (x + 1)^{m+1} = x^{m+1} + (m + 1)x^{m} + \frac{1}{2}(m + 1)m x^{m-1} + \etc.

  The binomial expansion of (x+1) raised to the power m+1, with its leading terms written out.

• Limiting Ratios of Magnitudes that Increase Without Limit

  \frac{a}{pa + b} = \frac{1}{p + \dfrac{b}{a}}

  A fraction with numerator a and denominator pa + b equals 1 divided by p + b/a, so it is near 1/p when b/a is small.

• Limiting Ratios of Magnitudes that Increase Without Limit

  \frac{A + (a + a' + a'' + \etc.)}{B + p(a + a' + a'' + \etc.) + b + b' + b'' + \etc.}

  The summed fraction with given quantities A and B added still approaches 1/p, provided the summed numerators become large compared with A and B.

• Limiting Ratios of Magnitudes that Increase Without Limit

  \frac{(x + 1)^{3} + (x + 2)^{3} + \dots + (x + n)^{3}}{(x + 1)^{4} - x^{4}}

  The sum of the numerators of the cube fractions, divided by the telescoped denominator (x+1)^4 - x^4, equal to the summed fraction in the argument.

• Limiting Ratios of Magnitudes that Increase Without Limit

  \frac{1^{3} + 2^{3} + 3^{3} + \dots + x^{3} + (x + 1)^{3} + \dots + (x + n)^{3}}{(x + n)^{4}}

  Adding x^4 to the denominator and the sum of cubes 1^3 to x^3 to the numerator leaves the ratio within the same nearness of 1/4, so the ratio of the sum of cubes to x^4 tends to 1/4.

• Limiting Ratios of Magnitudes that Increase Without Limit

  x\, \dfrac{x - 1}{2} ÷ x^{2} = \dfrac{x - 1}{2x}

  The ratio of x(x-1)/2 to x^2 simplifies to (x-1)/(2x), whose limit as x increases without limit is 1/2.

Problems

No exercises in this chapter.