Maths2

Országok listájaHungaryBudapesti Corvinus EgyetemGazdálkodástudományi KarNemzetközi gazdálkodás (angol nyelven)Mathematics1JegyzetekMaths2

2.1. LIMIT OF FUNCTIONS, CONTINUITY

2.1 Limit of functions, continuity
Basic notions
In the subsequent chapter we study the limit of functions f : R R. Let x0 be a point (possibly equal to ±) for which there exists a sequence xn in the domain of f such that xn x0 .

Denition 2.1 The limit of the function f at the point x0 is said to be A (which
can be ±) and denoted by
xx0

lim f (x) = A

if for any sequence xn from the domain of f whenever xn x0 , xn = x0 , then f (xn ) A.

Theorem 2.1 If the functions f and g have limits at x0 and limxx0 f (x) = A and limxx0 g(x) = B then
· limxx0 (f ± g)(x) = A ± B, · limxx0 (f · g)(x) = A · B, · if B = 0 then limxx0 f (x) = g
A B,

· if A = 0 and B = 0 then limxx0 f (x) = ±. g

Example 2.1 Determine the limit
x2 - 4 . x2 x - 2 lim

This function is not dened for x = 2 but it is equal to x + 2 at any point x = 2. Therefore it is easily seen that
x2 - 4 =4. x2 x - 2 lim

Example 2.2 Consider the function f (x) = 1/x. This function is not dened at

x = 0. On the other hand, for any sequence xn > 0, xn 0 from the domain we have f (xn ) + while f (-xn ) -. Thus this function has no limit at x = 0, that is 1 lim x0 x does not exist.

Example 2.3 Consider the following limit:
2x4 - 5x3 + x - 8 x+ 8x3 - x2 + 12 lim

Dividing both the numerator and denominator by x3 we get the expression
2x - 5 + 1/x2 - 8/x3 . 8 - 1/x + 12/x3

Now for any sequence xn + the limit of the numerator is +, while the limit of the denominator equals 8, thus the fraction tends to +. It can be seen in a similar way that the limit of the fraction is -, if x -.
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2.2.

SQUEEZING THEOREM

Example 2.4 Show that
x+

lim ( 1 + x2 - x) = 0 .

Indeed,

1 1 + x2 + x and the expression on the right hand side approaches 0 if x +. 1 + x2 - x =

2.2

Squeezing theorem

Similarly as for sequences, the Squeezing theorem holds for functions as well.

Theorem 2.2 Let f , g and h functions such that for any x
f (x) g(x) h(x)

and furthermore, limxx0 f (x) = limxx0 h(x) = A. Then the limit of the function g at x0 exists, and lim g(x) = A .
xx0

sin x ! x This is an even function, therefore it is enough to consider positive values of x. A geometric interpretation shows that for all 0 < x < /2
x0

Example 2.5 Find the limit

lim

sin x < x < tan x .
Dividing by sin x it is obtained, that

x 1 < . sin x cos x Thus, it follows by the squeezing theorem that x lim =1 x0 sin x and then taking the reciprocal, it follows that 1<
x0

lim

sin x =1 x

2.3 One-sided limit
Denition 2.2 We say that the right-hand limit of f at the point x0 exists and is
equal to
xx0 +

lim f (x) = A

if for any sequence xn x0 , xn > x0 from the domain of f we have f (xn ) A. The left-hand limit is dened analogously.

Example 2.6 Consider the function:
2x + 1 x-2 It is easy to see that if xn approaches 2 from the right then f (xn ) +, while if xn approaches 2 from the left then f (xn ) -. Therefore f (x) =
x2-

lim f (x) = -

and
2

x2+

lim f (x) = + .

2.4. CONTINUITY

2.4 Continuity
Denition 2.3 We say that the function f is continuous at a point x0 of its domain
if
xx0

lim f (x) = f (x0 ) .

If f is not continuous at a point x0 of its domain, then it is said that the function has a discontinuity there.
Attention! Continuity is dened only at points of the domain of a function. For instance the function f (x) = 1/x is continuous at each point of its domain, that is at each x = 0. The point x0 = 0 is not in the domain of f , so we cannot speak of discontinuity. On the other hand, f cannot be dened at x0 = 0 so that it becomes continuous, as the limit of the function does not exist there. Functions obtained from continuous function by composition or by basic operations (addition, subtraction, multiplication, division ) are also continuous except if the denominator of the fraction equals zero.

Example 2.7 For instance, consider the following function:
f (x) =
1-cos x x2 1 2

if x = 0 if x = 0

It is clear that this function is continuous for all x = 0, furthermore
1 - cos x 1 - cos2 x = = 2 x (1 + cos x)x2 sin x x
2

·

1 . 1 + cos x

This shows that the limit of the function at 0 equals 1/2. Thus, this function is continuous on the whole real line.
We think of a continuous function as one whose graph can be drawn by an unbroken curve. This is expressed in Bolzano's theorem.

Theorem 2.3 (Bolzano) Let f be a continuous function on the interval [a, b], and

suppose that f (a) and f (b) have dierent signs. Then there exists a point c (a, b) such that f (c) = 0.

Example 2.8 Decide if the equation
2x5 - 18x4 + 3x3 + 20x - 13 = 0

has a real solution? The expression on the left side of the equation denes a continuous function f for which
x+

lim f (x) = +

and

x-

lim f (x) = - .

Therefore f is positive for suciently large values of x and takes negative values if x is small enough. Therefore, by the Bolzano-theorem the equation has at least one real solution.
The following property of continuous functions is of fundamental importance for extremum problems and optimization.

Theorem 2.4 (Weierstrass) Let f be a continuous function on the interval [a, b].
Then f takes its maximum and minimum on this interval.
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2.4. CONTINUITY

For example, the function

f (x) =

x 3-x

if 0 x 1 if 1 < x 2

does not reach its maximum value on the interval [0, 2], but you see, it is not continuous at 1.

Exercise 2.1 Find the limit of functions:
x-1 x-1 x x ; lim ; lim lim ; x0 x1 x0 x - 1 x-1 x x x2 - b2 xn - bn xn - bn x2 - 25 lim ; lim ; lim ; lim k . x-5 x + 5 xb x - b xb x - b xb x - bk
x1

lim

x3 - x2 - x + 1 ; x1 x3 + x - 2 lim lim

x3 + x - 2 . x1 x3 - x2 - x + 1 lim

3 4 1 1 - ; lim - 3 ; x1- x - 1 x - 1 x3 - 1 x -1 4 4 1 1 lim - 3 ; lim - 3 ; x1+ x - 1 x1 x - 1 x -1 x -1 a b lim - 3 . x1 x - 1 x -1 3 3 x-1 x+1 ; lim . lim 5 5 x-1 x1 x-1 x+1 n 1+x-1 1-x-1 1+x-1 lim ; lim ; lim . x0 x0 x0 x x x
x1

The limit of a rational function

p(x) at ± depends of the degree of p(x) and q(x) if if if deg(p(x)) = deg(q(x)) = n deg(p(x)) > deg(q(x)) deg(p(x)) < deg(q(x))

q(x)

an bn p(x) lim = sgn(an ) ± x± q(x) 0

In the following example we show how the previous result is obtained.

Example 2.9
2 x2 (2 - + 2x2 - 3x + 5 x a lim 2 = lim 4 x 3x + 4x - 3 x 2 x (3 + - x 5 ) 2- x2 = lim 3 x ) 3+ x2 2 5 + x x2 = 2 . 3 4 3 - x x2

5 5 x(2x - 2 + ) 2x - 2 + 2x2 - 3x + 5 x = lim x = . b lim = lim 3 3 x x x 4x - 3 x(4 - ) 4- x x 3 5 3 5 x2 (- + 2 ) - + 2 -3x + 5 0 x x x x = lim c lim 2 = lim = =0 4 3 4 3 x 2 x 3x + 4x - 3 x 3 x (3 + - 2 ) (3 + - 2 ) x x x x

4

2.4. CONTINUITY

Exercise 2.2 Find the indicated limits:
4x2 + 1 ; x 3x2 - 2x + 5 lim x2 + 1 ; x 3x - 1 lim x2 - 2x + 1 . x- x3 + 2002 lim

In the subsequent problems use the result
x0

lim

sin x =1 x

to determine the indicated limits!

Exercise 2.3
sin ax ; x0 x lim
x0

sin ax ; x0 sin bx lim
x0

tan ax ; x0 x lim
x0

tan ax ; x0 tan bx lim 1 - cos ax ; x2
1-cos x x2 x0 1 2

sin ax . x0 tan bx lim lim 1 - cos x . sin2 x

lim

1 - cos x ; x

lim

1 - cos x ; x2

lim

Use the results limx0
lim

1-cos x x

= 0 and limx0 lim

=

to calculate
2

1 + sin x - cos x ; x0 1 - sin x - cos x

tan x - sin x ; x0 x3

x0

lim x

(1 - cos x) . tan3 x - sin3 x

The identities cos - x = sin x, sin - x = cos x and sin ( - x) = sin x 2 2 can be used to nd the limit in the following problems:
x 2

lim

- x tan x; 2

x

lim

sin x ; -x

x

lim

sin x . x2 1 - 2

If limxa f (x) = ±, then

xa

lim

1+

f (x)

f

(x) = e

R.

Exercise 2.4
limx 1 - limx 1 -
1 x x
2

;

limx

2x-1 2x+1
1

x

;

limx

x+1 2x-1

x

; ;

1 x x

;

limx0 (1 + x) x ;

limx0 (1 + tan x)

cot x

5



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