Analysis period 1.3 Flashcards
Limit of functions when x tends to infinity
Let f be a function with domain from reeel numbers, which contain large values. This function f has a limit as x approaches plus infinity, if and only if there exists L s.t for all eps >0 there exists M>0: |f(x) - L| < eps for all x>=M, x is an element of D
Limit of functions when x tends to minus infinity
Let f be a function with domain from reeel numbers, which contain large negative values. This function f has a limit as x approaches minus infinity, if and only if there exists L s.t for all eps >0 there exists M>0: |f(x) - L| < eps for all -x>=M, x is an element of D
Proving f has limit L when x tends to -+ infinity
- Always start with let epsilon >0 be given
- Start with |f(x) -L| and get rid of the absolute values. Replace x with -x if the infinity is negative
- Next, you make the term smaller than any other bigger term,, If you can get the latter term smaller than epsilon, then the term you began with is also smaller than epsilon. Note: It could be that your inequality is only valid upward of a certain point.
- Let M be the maximum of the values for which your inequalitis hold and the function of epsilon
- State the standard sentence and draw your conclusion
Horizontale asymptote
If lim x-> inf f(x) = L then the line y=L is called a Horizontale asymptote for the function f
limit of monotone functions
If the function f is defined on an unbounded above domain D C R and is eventually monotone and eventually bounded then lim x-> inf f(x) is finite
limit of function f which goes to +- infinity
Let f be a function with domain D c- R, which contains large (negative) values. We say that f tends to infinity as (-)x tends to infinity if and only if for all K>0 there exists M>0: f(x) > K provided that (-)x >= M and x is an element of D. lim f(x) = infinity
Let f be the function with domain D c- R which contains large values. The function f tends to -infinity as x tends to (-)infinity if and only if for all K>0 there exists M>0 -f(x)>K provided that (-)x >= M and x is an element of D. lim f(x) = minus infinity
Limit of functions when x tends to a is an element of R
Suppose that a function f:D -> R with D c R and suppose that a is an accumulation point of D. The function f has a limit as x approaches a if and only if there exists L such that for all eps >0 there exists delta > 0 |f(x) - L| < eps provided that |x - a| < delta and x is an element of D
lim x-> a f(x) = L
Limit of polynomials
Suppose that a function f is defined by f(x) = p(x) / q(x)
then:
1. If n<m then lim f(x) = 0
2. if n=m then lim f(x) = an / bn
3. if n > m then lim f(x) if infinite
Proving f goes to +- infinity when x tends to +- infinity
- Let K>0 be given
- Start with (-)f(x). Optional try to replace x in the function by -x when x -> infinity
- Next, you can make the term lager than any other, smaller term. If you can get the latter term larger than K, then the term you began with is also larger than K. Note: It could be that your inequality is only valid upward of a certain point
- Let M be the maximum of the values for which your inequalities hold and the function of K.
- State that there exists M>0 (-)f(x) > K for all (-)x >= K and draw your conclusion
Proving f has limit L element of R when x tends to a element of R
- let eps > 0 be given
- Start with |f(x) - L|
- You can rewrite |f(x) -L| to c x |x - a| ith constant c, so you can replace |x - a| with delta.
- next, if you solve some expression of delta for epsilon, you can choose your delta.
- State that there exists delta > 0 s.t |f(x) - L| < eps provided that |x - a| < delta and draw your conclusion
Properties limits when x tends to a ∈ R
Suppose that functions f,g,h : D -> R with D c- R, a is an accumulation point of D, and lim x-> a f(x) = A, lim x-> a g(x) = B, and lim x-> a h(x) = C. Then all of the conclusions for the previous properties theorem are true with infinity replaced by a and with eventually replaced by near x = a
Limit of a function and a sequence
Let the function f be defined on some deleted nieghborhood D of the real number a. The following two statements are equivalent
1. lim x-> a f(x) = L
2. For every sequence {xn} converging to x = a with xn ∈ D and xn =/ a eventually, the sequence {f(xn)} converges to L
Limit of functions to +- infinity when x tends to a ∈ R
Suppose that the function f : D -> R with D c R and a an accumulation point of D. Then the function f tends to (-) infinity as x approaches a if and only if for all K>0 there exists delta>0: (-)f(x) > K provided tht 0 < |x - a| < delta and x ∈ D
Whenever this is the case,
we write lim x->a f(x) = (-) inf
Product of the functions with infinite limits
Let the functions f and g be defined on some deleted neighborhood of x = a. If lim x->a f(x) = L > 0 and
lim x->a g(x) = + inf > 0 then lim x->a (fg)(x) = +infinity
Limit of sin(x)
let lim x->0 g(x) = 0. then lim x->0 sin(g(x))/ g(x) = 1
and lim x->0 g(x)/sin(g(x)) = 1
Side limit to L ∈ R
for the right hand limit
there exists L ∈ R s.t for all eps > 0 there exists delta > 0 |f(x) - L| < eps provided that 0 < x - a < delta and x∈D lim x-> a+ f(x) = L
for the left hand limit
there exists L ∈ R s.t for all eps > 0 there exists delta > 0 |f(x) - L| < eps provided that 0 < a - x < delta and x∈D lim x-> a- f(x) = L
Side limits to infinity
right
For all K>0 there exists delta >0: f(x) > K provided that 0< x-a < detla and x ∈ D lim x-> a+ f(x) = infinity
left
For all K>0 there exists delta >0: f(x) > K provided that 0< a - x < detla and x ∈ D lim x-> a- f(x) = infinity
Even and odd vertical asymptotes
If the limit from the right or from the left at x=a of a function f is infinite, meaning +- infinity then the line x=a is called a vertical asymptote
1. if both limits are infinite ( +infinite or -infinite) and equal, then asymptote x=a is even
2. If both limits are infinite (+infinity or -infinity) but opposite signs, then asymptote x=a is odd
3. If one of these limits is infinite and the other is finite or does not exist, then x=a is still a vertical asymptote, but it is neither odd nor even
Limits that do no ecist
The limit lim x->a f(x) exists if and only if both side limits tend to the same value. So lim x->a f(x) = L if and only if lim x->a- f(x) = lim x-> a+ f(x) = L
Limits of inverse arguments
Let a function f be defined for x∈(0,a) with a>0 a real number. If either of the limits lim x-> 0+ f(x) or lim t-> infinity f(1/t) exists, then both limit exists and are equal.