CHAPTER XIV
DIFFERENTIAL CALCULUS. MISCELLANEOUS
APPLICATIONS
LIMITING VALUES OR UNDETERMINED FORMS
- 1.In certain cases the value of a function cannot be at once
x"1ascertained. A common form is illustrated by Lt— , the value
x—1
of which is not apparent since it takes the form - . The function in this case is said to take an Undetermined or Indeterminate Form, and since, in the limit, its value is represented by the ratio of twoindefinitely small quantities, it may be expected that the methods of the Differential Calculus will enable the Limiting Value of the function to be found.
2.The process to be employed can best be illustrated by examples.
x'a — 1Example 1. To findLt ,I x—1Let x =1 + h, then.x" 1 =(1+h)"—1
zL1[x—1] hL0 (1+h)—1][n1+()h2+()hs+...] Lt h->oItL —'Lt[n+(72)h+(3)h'+...]= n.Example 2. To find Ltx—[log(1 + x)~ o ez —
xz x' x-- +—...
Lt log (1 + x) _ Lt2r—u-1z~u x+-+-+...2!3!= Lt [1— x + 2 x' — ...] (by actual division)xyo3= 1,since the series obtained by division is obviously convergent.
MISCELLANEOUS APPLICATIONS 83
Example 3. To find Lt gcz — g zyogz
Lt gez — g = Lt .g [gcx-' — 1]
.-0 gz — z—o gz — 1
z
= Lt g [gx log c+ 2 (log c)2+... — 1]
gz
z~.o [{x1ogc+(logc)2+ g ...} log g
+xloge+2i(log C)2+...~' (logg)2+...]
= Lt 11
z~u x log g + 2 i (log g)2 + ...
= Lt qlog,glogc+terms involving x2 and higher powers of x
x—.0 x log g + T, (log g)2 + ...
= Lt q logg log c+ terms involving x and higher powers of x
xyu logg+2i(log g)2+...
glogc.
3. The formulas of the Differential Calculus can be used for the solution of problems of this class, in the following way.
f Let ~~ take the form 0
when x--,-a.
Then if x = a + h, by Taylor's Theorem,
f (x) f (a) + hf' (a) + ...
(x) (a) +14' (a) + ... '
Now when x a, h -9.0 and both f (a) and (a) = O. We have, therefore,
Lt ' (x) _ f' (a)
xya4 (x) Y' (a),
If this is still an undetermined form, the process can be repeated. Taking Example 3 above, we have
| f(x)=g`z—g, |
~(x)=gz— 1, |
| f' (x) = g°z c'' log g log c, |
4' (x) = gz log g. |
6—S
84 DIFFERENTIAL CALCULUS
Therefore Lt~ fix) = Ltgc2 cx log g g log c g
=g log c, as before.
- Other Undetermined Forms are found when a function in the limit assumes one of the following forms :
co0 x co , — , cc — oo , 0°, co °, or 1°°.opThese can all be made to depend upon the form g .Thus if f (x) and j (x) be the two functions involved, in the first case by writing f (x) x (x) as ( ~~ _1 we have the form 0 and canproceed as before.Similarly for the second form.The third form can be written as co {1 — ~o ; the second term00within the bracket can then be evaluated, and the quantity within the bracket will then be found to be zero or otherwise. In the former event the product is of the form co x 0 and can be evaluated; in the latter event the value of the product is clearly infinite.The last three cases are solved by taking logs, when the logarithm of the function will be found to fall within one of the cases already discussed.
- In connection with the form 1°°, it is useful to remember that, from the development of the Exponential Theorem in algebra, it follows that
Lt(1+ =ex00xandLt (1 + a)x = ea.xymMAXIMA AND MINIMA
- If f (x) continuously increases with x until it attains a certain value a and subsequently decreases, the value a is said to be a maximum value of the function. Conversely if the value of the function continuously decreases, while x increases, until it attains a value b and subsequently increases, the value b is said to be a minimum value.
MISCELLANEOUS APPLICATIONS 85
By this definition there may be several "maximum" values and several "minimum" values; further, a "minimum" value may conbe greater than one of the "maximum" values.
These conceptions are illustrated in the following diagram.
Thus the points A and D are maxima, whilst the points B and E are minima. At the point C
the value of the function ceases D
momentarily to increase but is in-creasing both before and after the
point is reached. Such points, yE. where the rate of increase or de
crease in the value of the function tends to become constant as the point is reached, are called points of inflexion.
- 7.In obtaining criteria for ascertaining the maximum or minimum values the following considerations are of importance. If f (x) increases in value with x, it is clear from the definition
that dx is positive. Conversely if f (x) decreases with x, isnegative.
Further, if f (x) is continuous and assumes two values f (a) and f (b), one of which is positive and the other negative, it follows that for some value of x between the values a and b, ,f (x) must be zero.
- Now for a maximum value the function increases up to that value and then begins to decrease. Therefore at a maximum value
dx must change from positive to negative ; similarly at a minimum value is changing from negative to positive. It follows from the preceding article that at a maximum or minimum point the value of dx is zero.
- Further, at a maximum point dx is decreasing from positive to negative, and therefore dx2 must be negative or zero. At a minimum point the converse is the case and 2 is positive or zero.
86 DIFFERENTIAL CALCULUS
Hence in order to find the maximum and minimum values of f (x) it is necessary
- (i)to find the values of x which satisfy the equation f' (x) = 0,
to ascertain for each of these values whether, f" (x) is negative or positive.
If f" (x) = 0, a further test is necessary. For the investigation of these cases, which include values which give rise to points of inflexion, the student is referred to more advanced treatises on the subject.
Example 1. What fraction exceeds its pth power by the greatest number possible ?Let x be the fraction; we have to find the maximum value ofy=x—x",dx 1 — pxP-1.For a maximum or minimum value dx = 0, therefore 1—pxP-1=0,andxP-1= 1 p
or x =p Now x=—p(p—1)xP-2
P-2 = —p (p -1) p_
and is therefore negative.
Hence the above value of x is a maximum value.
Example 2. What are the dimensions of the largest rectangular box on a square base the area of whose surface does not exceed 12 square feet ?
Let x be the volume of the box, a the length of the base and b the height.
Then the surface is 2a2 + 4ab = 12, whence
, 6—a2 2a
MISCELLANEOUS APPLICATIONS 87
Also x = a' b
(6 —
2a ,
dx 3a'
da=3—2,
d'x=_3a.
da'
For a maximum or minimum value dx = 0, therefore
2
3—`2 =0
or a = V2, whence b = V2,
d'x and since (la'
is negative when a = V2, this value gives a maximum
value for x.