For instance, select the number 1000; we can take x so small 1 1 25.25 1 32.32 In the same way, x can be so chosen, that exceeds 10,000, or any other fixed number. х Hence the theorem, As the variable x is made to approach the limit zero, without α reaching zero, increases so as to exceed any large number that may be previously assigned. This theorem is sometimes condensed in this manner : As x approaches zero, α approaches infinity. A still further compression of the theorem consists in the use of the symbolism α =∞. This symbolism is objectionable, because it may convey the idea that division by 0 is permissible. The reader must remember that these symbols are intended merely as a convenient abbreviation for the theorem given above. a and b are fixed numbers (b0) and x is a variable approach ing the limit zero? Under these conditions the numerator a + x approaches a, the denominator b x approaches b. The reasoning is easier, if we first divide numerator and denominator by x, so that x occurs in the denominators of the minor fractions. α х Since a is a fixed number, and r increases without limit, the fraction approaches the limit zero, hence, the numerator 1 approaches 1 as a limit; the denominator +1 approaches + 1 as a limit; the com α α What is the limiting value of the following fractions when x approaches zero as a limit? What is the limiting value of the following fractions when the variable x increases without limit? SUPPLEMENT THE HIGHEST COMMON FACTOR BY THE METHOD OF DIVISION 189. The method of finding the h. c. f. of two polynomials not easily factored is similar to the method for finding the h. c. f. of two numbers in arithmetic, neither of which can be easily factored. The method was first suggested by Euclid (300 B.C.). Take the remainder 316 as a new divisor, and 1343 as a new dividend. Take the next remainder 79 as a new divisor, and 316 as a new dividend. The next remainder is zero. The last divisor 79 is the h. c. f. of 1343 and 3002. From the last equation we see that any factor of both A and B must be also a factor of B- AQ, and therefore a factor of the remainder R. Hence, all common factors of A and B are common factors of A and R. Instead of finding the h. c. f. of A and B, we may therefore find the h. c. f. of A and R; the latter process involves smaller numbers. Repeating the above process, divide A by R. Let Q' be the new quotient 4, and R' be the new remainder 79. As before, all common factors of A and R are factors of R'; for, A = Q'R + R' and R' = A — Q'R. N Hence, the problem reduces itself further, to finding the h. c. f. of R and R'; that is, to finding the h. c. f. of 316 and 79. As 316 = 79 × 4, it follows that 79 is the h. c. f. required. This process may be used in finding the h. c. f. of polynomials which cannot be easily factored. Numerical factors common to the polynomials are easily detected by inspection. Such numerical common factors, as well as other common factors that can be found by inspection, should be removed at once. EXAMPLE Find the h. c. f. of 2 x3-22x-12 and 4x3 +16 x2+16x+12. Process: 2 x3-22 x 12 = 2(x3 — 11 x — 6). 4x816x216x + 12 = 4(x3 + 4 x2 + 4 x + 3). The numerical factor 2 is common to the polynomials. To avoid fractions any expression may be multiplied or divided by any number which is not a factor of the other. In the example above, when 4 x2 in the first remainder will not be contained in x3 a whole number of times, the expression x3- 11 x - 6 is multiplied by 4. Again when 4x2 is not contained in 15 x2 a whole number of times, the - expression 15x2 53 x 24 is multiplied by 4. In the remainder 13x + 39, the factor 13 may be discarded because the expression 4x2+15x+9 does not have a factor 13. It would be wrong to multiply 4x2+15x+9 by 13, because 13 is a factor of 13 x + 39. Continue to divide until the remainder is of a lower degree than the divisor. Then take the remainder for a new divisor and the previous divisor for a new dividend and proceed as before. The 1. c. m. of two expressions not easily factored may be found by first finding their h. c. f. The 1. c. m. required will then be the product of either expression and the quotient obtained by dividing the other expression by the h. c. f. It is seldom that resort to this long process becomes necessary. 3. 12 y 30 y3 - 72 y2 and 32 y3+ 84 y2 - 176 y. 4. a4 a3 + 2 a2 - a + 1 and a1 + a3 + 2 a2 + a +1. 5. 8 am3 +24 am2 + 22 am +6 a and 6 m3 +13 m2 + 8 m + 1. 6. 2x3- 10 x2 - 14x + 70 and x3-3x2 — 7 x — 15. 7. 8x1y-10 x3y+7x2y-2 xy and 6xy-11 x1y+8 x3y — 2 x2y. 8. 4a3 4 a2 5 a 3 and 10 a2 - 19 a + 6. The h. c. f. of three or more expressions may be found as follows: Find the h. c. f. of two of them; then find the h. c. f. of this result and another of the given expressions, and so on. The last h. c. f. is the one required. 9. 2x3-2x2 - 2 x − 4, 3 x3 — 6 x2+9x-18, and 8x3-12x2 - 4x-8. 10. a3x-6a2x+11ax-6x, a3x-9 a2x+26 ax-24 x, and a3x2 - 8 a2x2 + 19 ax2 - 12 x2. |