..Definitions
and properties
..Special
techniques to calculate lim f(x)
..Continuity
Definitions and
properties
Definition
Say f: R -> R : x -> f(x) is a real function and a value b.If b is a real number, assume that f(x) is defined in ]b-e,b+e[ or ]b-e,b[ or ]b,b+e[ .
If b is +infinity, assume that f(x) is defined for all x > a fix number N.
If b is -infinity, assume that f(x) is defined for all x < a fix number N.
Take the set of all sequences {xn} such that
lim xn = b ;
for each n, xn is different from b ;
for each n, xn is in the domain of f.
With each sequence {xn} corresponds an 'image sequence'
{f(xn)}.If, for all these image sequences, lim f(xn) = (a fixed value c), then we say that lim f(x) = c . We write
lim f(x) = c or lim f(x) = c
x->b b
If there isn't such value c, we say that lim f(x) is not defined.
Examples
x.x - 5x + 6 (x - 2)(x - 3)
lim --------------- = lim --------------- = 1
3 x - 3 3 x - 3
lim sqrt(x) is not defined
-2
2.x2 + x 2
lim ------------ = ---
+infty 3.x2 + 4 3
Properties
If both sides exist, it can be proved that :
lim c.f(x) = c . lim f(x) (with c = constant)
b b
lim |f(x)|= |lim f(x)|
b b
p p
lim f(x) = (lim f(x) )
b b
lim pth-root(f(x)) = pth-root(lim f(x) )
b b
lim (f(x) + g(x)) = lim f(x) + lim g(x)
b b b
lim (f(x) . g(x)) = lim f(x) . lim g(x)
b b b
lim (f(x) / g(x)) = lim f(x) / lim g(x)
b b b
Some special limits
If both sides exist, we have
lim r = r (for each constant number r)
b
lim x = b
b
n n
lim x = b (n is positive integer)
b
lim (1/x) = 0
infty
n
lim (1/x) = 0 (n is positive integer)
infty
lim nth-root(x) = +infty
+infty
lim nth-root(x) = -infty
-infty
lim 1/nth-root(x) = 0
+infty
If T(x) and N(x) are polynomials
lim T(x) = T(b)
b
T(x) T(b)
lim ---- = ----
b N(x) N(b)
lim sqrt(T(x)) = sqrt(T(b))
b
Left and right limit
Say f: R -> R : x -> f(x) is a real function and there is a strictly positive real number e such that ]b-e,b[ is part of the domain of the function f.We restrict the domain of f to ]b-e,b[.
With this new domain, we take
lim f(x)
b
This limit is called 'the left limit of f(x) in b' .This left limit is noted
lim f(x)
< b
Analogous: f: R -> R : x -> f(x) is a real function and there is a
strictly positive real number e such that ]b,b+e[ is part of the domain of the
function f.We restrict the domain of f to ]b,b+e[.
With this new domain, we take
lim f(x)
b
This limit is called 'the right limit of f(x) in b' .This right limit is noted
lim f(x)
> b
Important example
1
lim ------ = - infty
< b x - b
1
lim ------ = + infty
> b x - b
Corollary
If lim f(x) not = lim f(x) then lim f(x) is not defined < b > b bIf lim f(x) = lim f(x) = c then lim f(x) = c < b > b b
If lim f(x) = c then we have not always lim f(x) = lim f(x) = c
b < b > b
Special techniques to calculate lim f(x)
The case +infty - infty with polynomials
Example
lim (2x2 - 6x + 7) =
infty
lim 2x2 .(1 - 3/(2x) + 7/(2x.x) ) =
infty
lim 2x2 . lim (1 - 3/(2x) + 7/(2x.x) ) =
infty infty
lim 2x2
infty
Rule:
If T(x) = a.xn + ... + l is a polynomial then
lim T(x) = lim (a.xn )
infty infty
The case infty / infty with a quotient of polynomials
In the same way as above you can prove that
If T(x) = a.xn + ... + l is a polynomial then
And N(x) = b.xm + ... + k is a polynomial then
T(x) ( a.xn )
lim ----- = lim -------
infty N(x) infty ( b.xm )
k/0 with a quotient of polynomials
Example:
-5x - 81 -5x - 81 1
lim -------------- = lim --------- . -------
> -3 (x + 3)(x - 1) > -3 (x - 1) (x + 3)
= (16.5) . (+infty) = (+infty)
----------
-5x - 81 -5x - 81 1
lim -------------- = lim --------- . -------
< -3 (x + 3)(x - 1) < -3 (x - 1) (x + 3)
= (16.5) . (-infty) = (-infty)
0/0 with a quotient of polynomials
Example 1:
2x.x - 4x 2x.(x - 2)
lim -------------- = lim ---------------
> 2 x.x - 4x + 4 > 2 (x - 2)(x - 2)
2x
= lim --------- = +infty
> 2 (x - 2)
Example 2:
2x.x - 4x 2x.(x - 2)
lim -------------- = lim ---------------
2 x.x - 5x + 6 2 (x - 2)(x - 3)
2x
= lim --------- = -4
2 (x - 3)
0/0 with irrational functions
Example 1:
sqrt(x-3) -1
lim ------------- =
4 x - 4
(sqrt(x-3) -1)(sqrt(x-3) +1)
lim ---------------------------- =
4 (x - 4) (sqrt(x-3) +1)
(x - 3 - 1)
lim ---------------------------- =
4 (x - 4) (sqrt(x-3) +1)
1
lim ---------------- = 0.5
4 (sqrt(x-3) +1)
Example 2:
______________
| 2
1 - \| x - 3 x + 3
lim ---------------------
1 _________
| 2
\| 4 x - 3 - 1
We multiply the nominator and the denominator with the factor F =
_____________ _________
| 2 | 2
( 1 + \| x - 3 x + 3)( \| 4 x - 3 + 1 )
_________
2 | 2
(1 - x + 3 x - 3) ( \| 4 x - 3 + 1 )
= lim --------------------------------------------------------
1 _____________
2 | 2
( 4 x - 4 ) ( 1 + \| x - 3 x + 3)
(1 - x2 + 3 x - 3)
= lim ---------------------- = ... = 1/8
1 ( 4 x2 - 4 )
Example 3
______________ _________
3| 3 3| 2
\| x - 2 x - 3 - \| 2 x - 7
lim ----------------------------------
2 2 x3 + x - 18
We multiply the nominator and the denominator with the factor F =
_________________ ___________________________ _____________
3| 3 2 3| 3 2 3| 2 2
\| (x - 2 x - 3) + \| (x - 2 x - 3) (2 x - 7) + \| (2 x - 7)
Then we have for the limit
x3 - 2 x - 3 - 2 x2 + 7
lim ----------------------------
2 (2 x3 + x - 18) . F
(x2 - 2) (x - 2)
=lim ----------------------------
2 (2 x2 + 4 x + 9) (x - 2) . F
(x2 - 2) 2
=lim ----------------------- = --------
2 (2 x2 + 4 x + 9) . F 25 .3
k/0 with irrational functions
Example 1:
1 + sqrt(-x)
lim -------------- =
> -2 x + 2
(1 + sqrt(-x)) 1
lim ---------------.-------- = +infty
>-2 1 (x + 2)
Example 2:
x2 - 5x + 4
lim ------------------- =
>3 sqrt(x2 - 5x + 6)
1
lim (x2 - 5x + 4).------------------- =
>3 sqrt( (x-2)(x-3) )
(-2).(+infty) = -infty
Example 3:
x2 - 5 x + 4
lim ----------------- =
<3 _____________
| 2
\| x - 5 x + 6
1
lim (x2 - 5x + 4).------------------- =
<3 sqrt( (x-2)(x-3) )
is not defined
infty/infty with irrational functions
Example 1:
________
| 2
\| x + 1 + 3 x
lim ------------------- =
+infty 2x - 5
_________
| -2
(\| 1 + x + 3) x
lim ----------------------- =
+infty x.( 2 - 5/x)
_________
| -2
(\| 1 + x + 3) 4
lim -------------------------- = --- = 2
+infty ( 2 - 5/x) 2
Example 2:
________
| 2
\| x + 1 + 3 x
lim ------------------- =
-infty 2x - 5
_________
| -2
x(3 - \| 1 + x )
lim ------------------------ =
-infty x.( 2 - 5/x)
_________
| -2
(3 - \| 1 + x ) 2
lim -------------------------- = --- = 1
-infty ( 2 - 5/x) 2
+infty - infty with irrational functions
Example 1:
________________
| 2
lim ( \| 4 x + 3 x - 1 + 2 x ) =
-infty
________________ ________________
| 2 | 2
(\| 4 x + 3 x - 1 + 2 x)(\| 4 x + 3 x - 1 - 2 x)
lim ----------------------------------------------------- =
-infty ________________
| 2
(\| 4 x + 3 x - 1 - 2 x)
(4x2 + 3x - 1) - 4x2
lim ------------------------------------- =
-infty ________________
| 2
(\| 4 x + 3 x - 1 - 2 x)
( 3x - 1)
lim -------------------------------- =
-infty ________________
| 2
(\| 4 x + 3 x - 1 - 2 x)
x ( 3 - 1/x)
lim -------------------------------- =
-infty _____________
| 3 -2
(- | 4 + - - x - 2) x
\| x
( 3 - 1/x) 3
lim -------------------------------- = ----
-infty _____________ 4
| 3 -2
(- | 4 + - - x - 2)
\| x
Example 2:
_____________
| 2
lim ( 5 + \| 4 x - x + 3 + 2 x )
-infty
_____________
| 2
= lim (2 x + \| 4 x - x + 3 ) + 5
-infty
_____________ _____________
| 2 | 2
(2 x + \| 4 x - x + 3 )(2 x - \| 4 x - x + 3 )
= lim --------------------------------------------------- + 5
-infty _____________
| 2
(2 x - \| 4 x - x + 3 )
x - 3
= lim ------------------------------ + 5
-infty _____________
| 2
(2 x - \| 4 x - x + 3 )
x( 1 - 3/x )
= lim ------------------------------ + 5
-infty _________________
| 2
x (2 + \| 4 - 1/x + 3/x )
( 1 - 3/x )
= lim ------------------------------ + 5 = 1/4 + 5
-infty _________________
| 2
(2 + \| 4 - 1/x + 3/x )
Many other special limits
For calculation of many other special limits (and examples) using derivatives seeTwo special limits and L'Hospitals rule and Examples .
Continuity
Definitions
Let f(x) be a real function.f(x) is continuous in b <=> lim f(x) = f(b) bf(x) is left continuous in b <=> lim f(x) = f(b) < bf(x) is right continuous in b <=> lim f(x) = f(b) > b
Collary
If f(x) is left continuous and right continuous in b, then f(x) is continuous in bIs f+g continuous?
If f and g are continuous in b, then f+g is continuous in b. Prove:
f is continuous in b => lim f(x) = f(b)
b
g is continuous in b => lim g(x) = g(b)
b
So,
lim (f(x) + g(x)) = lim f(x) + lim g(x) = f(b) + g(b)
b b b
Q.E.D. ..In the same way it you can prove that : If f and g are continuous in b, then f-g is continuous in b. If f and g are continuous in b, then f.g is continuous in b. If f and g are continuous in b, then f/g is continuous in b. (with g(b) not 0)
Theorem
It can be proved that all algebraic and trigonometric functions are continuous in all elements of their domain.Continuity of f(x) in an interval.
f(x) is continuous in an interval [a,b]
<=>
f(x) is continuous in each element of [a,b]
Bolzano's theorem
If f(x) is continuous in [a,b] and f(a).f(b) < 0Then there is a real number c in ]a,b[ such that f(c) = 0.
Prove:
We'll prove the theorem for f(a) < 0 and f(b) > 0.
- Take m = (a + b)/2 . If f(m) = 0 , the theorem is proved.
In the other case, there is an interval [a1,b1], with a1 = m or b1 = m, such that f(a1) < 0 and f(b1) > 0. - Again, take m1 = (a1 + b1)/2 . If
f(m1) = 0 , the theorem is proved. .... After n steps we have
[an,bn] with f(an) < 0 and
f(bn) > 0.
The length of [an,bn] is(b-a) ------ 2n - In that way we have the sequences
a =< a1 =< a2 =< a3 =< a4 ...
b >= b1 >= b2 >= b3 >= b4 ...
The first sequence is not descending and has an upper bound. So lim an = c1.
The second sequence is not rising and has a lower bound. So lim bn = c2.
But,(b-a) lim (bn-an) = lim ----- = 0 => lim an = lim bn =c1 = c2 = c infty infty 2n Now, Since f(x) is continuous in c , lim f(x) = f(c) cHence, for each sequence {xn} with limit c, the sequence {f(xn)} has limit f(c). Thus,lim f(an) = f(c) = lim f(bn) infty inftyAll terms f(an) are < 0 , thus f(c) =< 0 .
All terms f(bn) are > 0 , thus f(c) >= 0 .
So, f(c) must be 0.
Theorem of the mean values.
If f(x) is continuous in [a,b] and r is a real number between f(a) and f(b), then there is a number c in ]a,b[ such that f(c) = r.Prove:
Construct the function g(x) = f(x) - r.
Since f(x) and r are both continuous in [a,b], g(x) is continuous in [a,b].
Since r is a real number between f(a) and f(b), 0 is a real number between g(a) and g(b).
So, we can apply the Bolzano theorem on g(x) in [a,b].
Hence, there is a number c in ]a,b[ such that g(c) = 0.
This means that there is a number c in ]a,b[ such that f(c) = r.
Theorem of Weierstrass.
If f is continuous in [a,b] , then f(x) attains a maximal and a minimal image in [a,b]. For a proof of the theorem
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