- •Chapter 3. An introduction to calculus
- •§ 3.1. Functions
- •§ 3.2. Composite Functions
- •§ 3.3. Polar Coordinates
- •§ 3.4. The Theory of Limits
- •3.4.1. The limit of a sequence.
- •3.4.3. Infinitesimals and bounded functions.
- •3.4.4. The infinitesimals and their properties.
- •§ 3.5. Fundamental Theorems on Limits
- •§ 3.6. The First Remarkable Limit and Its Generalization
- •§ 3.7. The Second Remarkable Limit
- •§ 3.8. The Second Generalized Remarkable Limit
- •§ 3.9. Other Remarkable Limits
- •§ 3.10. Continuity of Functions
§ 3.5. Fundamental Theorems on Limits
Theorem I. The limit of the algebraic sum of finitely many functions equals the sum of the limits of these functions:
.
Proof. For simplicity, we shall prove the theorem for the sum of two functions. Suppose that . By Theorem I, these functions can be represented in the form u1(x)=b1+1(x), u2(x)=b2+2(x), where 1(х), 2(х) are infinitesimals. Consider the sum
u1(x)+u2(x)=b1 +b2+1(x)+2(x), where 1(x)+2(x) is infinitesimal.
Using the second part of Theorem I, we obtain
, as required.
Theorem II. The limit of the product of two functions equals the product of the limits of these functions:
.
Proof. Suppose that and . By Theorem I, we have u1(x)=b1+1(x) and u2(x)=b2+2(x).
Consider the product
u1(x).u2(x)=b1.b2+b12(x)+b21(x)+ 1(x)2(x)=b1.b2+(x),
where (х) is the infinitesimal sum of the last three terms. Theorem I implies , as required.
Theorem III. The limit of the ratio of two functions equals the quotient of the limits of the numerator and the denominator:
.
Proof. Suppose that
.
Then
u(x)=b1+1(x) and v(x)=b2+2(x).
Consider the quotient and transform it by reducing to a common denominator:
.
It is easy to show that the second quotient is infinitesimal. The fundamental Theorem I implies the required assertion.
Theorem IV (the sandwich theorem). If, in a neighborhood of a point а, the inequalities
u (x) z (x) v (x) (3)
hold , then limit of z(x) exists and equals .
Proof. Subtracting b from inequalities (3), we obtain (x)–b z(x)–bv(x)–b. By definitions, if |x–a|<, then |u(x)–b|< and |v(x)–b|<; hence –<u(x)–b<, and –<v(x)–b< . Therefore, for all |x–a|<, we have –<z(x)–b<, or |z(x)–b|<, which means that .
Theorem V. If a function f(x) is monotonically increases (decreases) and is bounded as х а above (below), then f(x) has a finite limit (Without proof.)
y
M
y=f(x)
b
0 a x
3.5.1. Computations of limits. Examples.
I. Limits as x.
(1)
The limits in the numerator and the denominator equal zero.
To find the limit of a linear-fractional function, we must divide the numerator and the denominator by х to the maximum power among the powers of x in the numerator and the denominator.
(2)
because х4 is the maximum power of x in the numerator and the denominator.
(3) (divide by х2).
A simple method for finding limits of linear-fractional functions as х is to leave the term containing the maximum power of х in the numerator and the denominator:
4) ,
5) ,
6) .
Let us find limits (1), (2), (3) by the simple method:
,
,
.
Deleting the terms containing lower powers of x from the numerator and the denominator is only possible because, after division by х to the maximum power, the limits of all such terms vanish.
II. Limits as ха. Looking for a limit, first, substitute in the function. If we obtain a number, then this number is the limit of the function. If we obtain one of the indeterminacies ,1, and , then we must eliminate it by transforming the function and then to pass to the limit.
(1) ,
(2) ,
(3) ,
(4)
.