Chapter 3: limits of functions Flashcards
Cartesian product A x b
Set of ordered pairs:
A x B = { (a,b) : a∈A ,b∈B} where (a,b) = {{a}, {a,b}}
A relation
A relation is a subset R of A x B:
aRb if (a,b)∈R
The domain of the relation is the set:
D_R = {a∈A; aRb for some b∈B}
The relation R is functional if for all a∈D_R there is exactly one b∈B so that aRb
(In this case b=f(a) instead of aRb and D_f instead of D_R.
f:A->B mapping with domain D_f and image ={b∈B, b=f(a) for some a ∈A}
If f and g are two functions from A to B with domains we can say f=g if
For functions f and g from A to B with domains D_f and D_g respectively we sat f=g if
D_f =D_g and f(x)=g(x) for all x ∈D_f
Domain D_f
For polynomial
For rational function
Domain D_f us all the points x∈R for which f(x)∈R. Any x∈R for which f(x) =0/0 or f(x) = plus or minus infinity is not in D_f
- for f(x) = a₀+a₁x+… + a_n-1x^(n-1) + a_nx^n with coefficients a₁,… ∈R and the degree n∈N, D_f =R
- f(x) = p(x)/q(x) where p,q polynomials D_f = {x∈R: q(x) not equal to 0}
Algebra of functions
For f,g : R -> R and α∈R
With D_f and D_g
Domains and functions of:
(f+g)(x)
(αf)(x)
(fg)(x)
(f/g)(x)
• (f+g)(x) =f(x) + g(x)
D_(f+g) = intersection if D_f and D_g
•(αf)(x) =αf(x)
D_αf = D_f
•(fg)(x)=f(x)g(x) D_fg = intersection again
•(f/g)(x)= f(x)/g(x) provided g(x) is not equal to 0 then domain D_(f/g) = intersection again \ { x in D_g : g(x) =0}
Composition makes sense if image is subset of domain
Definition: limit of a function and sequences
We say that f has a limit l at a point a in R if limit as n tends to infinity of f(x_n) =l for every sequence (x_n) which satisfies:
1) for all n in N, x_n in D_f{a}
2) limit as n tends to infinity of x_n =a.
If this holds we write
Limit as x tends to a of f(x) =l
Note: we must have convergence for every such sequence but we don’t need to have a in the domain. In fact if a isn’t in the domain D_f then D_f{a} = D_f
Thm 3.2.1: the algebra of limits LIMITS
Suppose that f,g:R->R and a∈R is such that limit as x tends to a of f(x) =l and limit as x tends to a of g(x) =m.
Then
1) limit as x tends to a of (f+g)(x) =l+m
2) limit as x tends to a of (fg)(x) =lm
3) limit as x tends to a of (αf)(x) =αl
4) limit as x tends to a of (f/g)(x) =l/m if m is not equal to 0
Proof:
These all follow from the ALGEBRA OF LIMITS FOR SEQUENCES THEOREM 2.2.1
1) if (x_n) is arbitrary sequence in (D_f ∩ D_g){a} that converges to a then
Limit as n tends to infinity of (f+g)(x_n)
= limit as n tends to infinity of (f(x_n) +g(x_n)) by the algebra of limits for sequences !!!
=limit as n tends to infinity of f(x_n) + limit as n tends to infinity of g(x_n)
=l+m by the algebra of limits
Similarly 2) (D_f ∩ D_g){a} (without a remember)
and 3) in D_f
With 4) (D_f ∩ D_g)[{a}∪{y:g(y)=0}]
(ε-δ) criterion for existence of limits
Theorem 3.2.2
Let f be a mapping from R to R, then the limit as x tends to a of f(x) =l
IF AND ONLY IF
The (ε-δ) criterion holds:
Ie given any ε>0 there exists δ>0 so that whenever x∈D_f with
0 less than |x-a| less than δ,
Then | f(x) -l | less than ε
Theorem 3.2.3:
Sandwich rule for functions
Suppose that f,g,h:R->R and suppose that there exists an interval (a,b)⊆ D_f ∩ D_g ∩ D_h so that for all x∈(a,b)
f(x) ≤ g(x) ≤ h(x)
If there exists c∈ (a,b) such that limit as x tends to x of f(x) = limit as x tends to x of h(x) =l, then limit as x tends to c of g(x) exists and equals l.
Proof:
Let (x_n) be any sequence that converges to c, by taking N∈N sufficiently large ensures that x_n ∈(a,b) for all n>N. then the result follows by the sandwich rule for sequences(THM 2.2.2)
By taking the sequences there to be a_n= f(x_(n+N)), b_n = g(x_(n+N)) and c_n=h(x_(n+N)) for all n∈N
Theorem3.3.1:
Theorem for (ε-δ) criterion left and right hand limits
Let f be a mapping from R to R , then:
1•limit as x→a+ [f(x)] =l₁
2•Lim_( x↑a) [f(x)] =l_2
IF AND ONLY IF
Given any ε>0 there exists δ>0 so that whenever x∈D_f with
1•0< x-a less than δ, then |f(x) -l₁| less than ε
2• 0< a-x less than δ, then |f(x) -l_2| less than ε
Theorem 3.3.2 both left and right limits and limit at a point
Relationship
A function f:R->R has a limit at a point x=a
IF AND ONLY IF
Both left and right limits EXIST at x=a and they are equal
In which case limit takes same common value
Proof: suppose f has a limit at x=a, then for every sequence (x_n) with x_n in
Example a rational function has domain st
St that denominator never 0 eg R{3,5}
What is the domain
D_f us all points in Reals for which f(x) is in R, ie any number for which value is 1/0 or 0/0 cannot be in the domain
Eg rational function omits values which are 0 in denominator function.
Eg the sign function has the reals as a domain
Eg indicator function has R also
The limit of a function can be found for a not in the domain. Ie when there is a break.
Example: investigating a point not in the domain
f(x) = (x-5) / (x^2 -25)(x-3)
This had domain D_f = R{-5,3,5}.
Observe that f(x) =(1)/(x+5)(x-3) for all x in the domain.
To investigate x=5 we choose an arbitrary sequence (x_n) with x_n in D_f for all with limit as n tends to infinity of x_n as 5.
A PARTICULAR example is 5 + (1/n) or 5 +(7/n^2) etc
Then f(x_n) = 1/(x_n +5)(x_n -3) and the limit as n tends to infinity is 1/(10 times 2) = 1/20.
Hence the limit of f(x) as x tends to 5 is 1/20.
THE LIMIT EXIATS EVEN THOUFH ITS NOT IN THE DOMAIN.
We can investigate x=-5 again in a similar way. Using a particular sequence. But this time we find that f(x) it diverges to -infinity. In this example we find it’s continuous at a as the limits exist and equal f(a) for points NOT -5,3 and 5. These aren’t in the domain.
