Topic 6: Limits Flashcards
(15 cards)
What is the informal definition of limits (2)
-Take function f: ℝ -> ℝ and some fixed value a ∈ ℝ
-If we can make f(x) arbitrarily close to some finite value L ∈ ℝ just by making x arbitrarily close to a, then we say that f(x) tends to L as x tends to a
What is the formal definition of limits (2)
-Let f(ℝ -> ℝ) be a function, and suppose a ∈ ℝ
-Then f(x) tends to the limit L when x tends to a if for any ε > 0, there exists some δ = δ(ε) > 0 (delta which typically depends on epsilon) such that if |x-a| < δ, we have |f(x) - L| < ε
What does it mean for a limit to be well defined (2)
-The limit exists and has an unique value
-This means that the left-hand limit and the right-hand limit are equal
What are the algebra of limits (3)
Let f: ℝ -> ℝ, g: ℝ -> ℝ, suppose f(x) -> L, g(x) -> m as x -> a
-kf(x) -> kL for any k ∈ ℝ
-f(x) + g(x) -> L+M as x -> a
-f(x)g(x) -> LM as x -> a
What is the sandwich rule for limits (3)
-Suppose f, g, h: ℝ -> ℝ are functions
-For some a ∈ ℝ we have lim f(x) = L, lim h(x) = L and f(x) ≤ g(x) ≤ h(x) over some open interval contained by a
-Then lim g(x) = L
How can we prove the sandwich rule for limits (2,3)
-Since lim f(x) = L, then for any ε > 0, there exists δ1 > 0 such that L - ε < f(x) < L + ε for all |x - a| < δ1
-Similarly, since lim h(x) = L, then for any ε > 0, there exists δ2 > 0 such that L - ε < f(x) < L + ε for all |x - a| < δ2
-Now choose δ = min{δ1, δ2}
-Then L - ε < f(x) ≤ g(x) ≤ h(x) < L + ε for |x - a| < δ
=> L - ε < g(x) < L + ε as well
How can we consider Limx -> 0 sin(x)/x (2,3)
-For 0 < x < π/2, we have sin(x) < x < tan(x) = sin(x)/cos(x)
=> 1 < x/sin(x) < 1/cos (x) (divide everything by sin(x))
=> cos x < sin(x)/x < 1
-Now define h(x) = 1, f(x) = cos(x), and g(x) = sin(x)/x
-as x ->0, we know the limit of h(x) and f(x) is 0
-Therefore, by the sandwich rule, the limit as x->0 of sinx/x = 1
How can we prove limx->a (xm - am)/x-a = mam-1 for all m ∈ ℤ ()
-Let m ∈ ℕ = ℤ+
-Observe xm - am = (x-a)(xm-1 + xm-2a + … + xam-2 + am-1)
-Then limx->a (xm - am)/(x-a) = limx->a</sub> (xm-1 + xm-2a + … + xam-2 + am-1)/(x-a)
= limx-a xm-1 + xm-2a + … + xam-2 + am-1
-Set x = a
-am-1 + am-1 + … (m terms)
=mami1
-If n is a negative integer, then we right m = -n for some n ∈ ℕ
-xm - am = x-n - a-n
=(an - xn)/(xnan
-Then, (xm - am)/x-a = (xn - an)/(xnan(x-a)
=-(nan-1)/a2n = -na-n-1 = mam-1
-Hence, the result holds for m ∈ ℤ
What does 1/x tend to as x -> 0 (3)
-Take f: ℝ+ -> ℝ given by f(x) = 1/x
-As x -> 0+, f(x) gets arbitrarily large (limits are for finite real numbers)
-Informally, we can say f(x) tends to +∞
What is some terminology referring to f(x) = (1/x)(sin(1/x)) (1,2,2)
-Let f:ℝ{0} -> ℝ by f(x) = (1/x)(sin(1/x))
-Sin(1/x) oscillates finitely between +-1 as x -> 0
-(1/x)sin(1/x) oscillates infinitely as x->0
-1/x, 1/xn, (1/x)sin(1/x) are all unbounded near x = 0
-sin(1/x) is bounded near x = 0
What is a proposition about f(x) being bounded near x = a (2)
-If f(x) -> L as x -> a, f(x) is bounded near x = a
-We can find an open interval (c, d) containing a, such that f(x) is bounded on (c,d)
How can we write limits when x -> +-∞ (2)
-If x is sufficiently large, 1/x is arbitrarily small
-Intuitively and informally, it makes sense to say 1/x ->0 as x->∞
What is the definition of a limit as x -> ∞ (3)
-Say f(x) can be made arbitrarily close to some finite value L ∈ ℝ just by making x sufficiently large
-Then we say f(x) tends to the limit L as x tends to ∞
-We write f(x) -> L as x -> ∞ or limx->∞ f(x) = L
What is a useful method for calculating limits as x -> ∞ (2)
-Write x = 1/t
-If x -> ∞, t -> 0+ (from the positive side)
How can we calculate limx->∞ (3x2 + 1)/(x2 - x + 1) (4)
-limx->∞ (3x2 + 1)/(x2 - x + 1)
-limt->0+ (3(1/t)2 + 1)/((1/t)2 - (1/t) + 1)
-limt->0+ (3 + t2)/(1 - t + t2) = 3
