%%visits: 2 Limit, it is used in [[differentiation]], in [[convergence]] [and a lot of other places] ## intuition A limit to infinity is pretty much [[convergence]] with different names. ∀ϵ > 0∃N ∈ ℕ∀n ≥ N : |an − a| < ϵ now make n = x N − R ℕ = ℝ and an = f(x), a = y0 ∀ϵ > 0∃N ∈ ℕ∀n ≥ N : |an − a| < ϵ
now a limit as x → ∞ is also similar, in the sense that instead of R we have a neighbourhood of x0, f(x) must be in a punctured neighbourhood of x0 but not necessarily at f(x0).
limit at a poin := ∀
neighbourhood U of y0, ∃ a neighbourhood V of x0 such that if x ∈ V, then f(x) ∈ U. limit at
infinit := ∀ neighbourhood
U of y0, ∃ a neighbourhood V of ∞ such that if x ∈ V, then f(x) ∈ U.
$$ \definecolor{c1}{RGB}{255, 0, 0} \definecolor{c2}{RGB}{0, 0, 255} \definecolor{c3}{RGB}{255, 165, 0} \definecolor{c4}{RGB}{75, 0, 130} \definecolor{c5}{RGB}{220, 220, 0} \definecolor{c6}{RGB}{238, 130, 238} \definecolor{c7}{RGB}{0, 128, 0} \definecolor{c8}{RGB}{100,100,100} \definecolor{c9}{RGB}{45,177,93} \definecolor{default}{RGB}{10,20,20} $$
$\color{default}$
$$ \lim_{ x \to x_0 } \left[ f(x) \right] # l $$ is just shorthand for $$ \color{c1} \left| f(x) - l \right| \color{c2}< \epsilon , \forall \epsilon > 0 $$ where 0 < |x − x0| < δ, ∀δ > 0 $\color{c1}$ the difference between f(x) and l $\color{c2}$ gets smaller and smaller $\color{c3}$ as long as the difference between x and x0 gets smaller and smaller
δ is dependent on ϵ, that means if you change epsilon, your value of delta should be different. Note that as it is a modulus, then naturally we begin to see that we consider points above and below L. We are making a rigorous definition of the converging phenomena Finding if a limit exists can be done with L’hopital’s rule, and differentiation has an implied limit.
limx → x0[f(x)g(x)] = limx → x0[f(x)]limx → x0[g(x)]
limx → x0[f(x) + g(x)] = limx → x0[f(x)] + limx → x0[g(x)] (BE CAREFUL OF ∞ − ∞)
$\lim_{x\rightarrow x_0}[\frac{f(x)}{g(x)}] = \frac{\lim_{x\rightarrow x_0}[f(x)]}{\lim_{x\rightarrow x_0}[g(x)]}$
Order matters when dealing with double limits Intuition regarding
cancelling “zero” terms When finding the limit $\lim_{x\rightarrow
2}[\frac{(2x+1)(x-2)}{x-2}]$ it makes the problem much easier
when we cancel x-2 from the top and bottom. BUT this is not the
same function. But what we are doing is we are saying x ≠ 2 so we cancel the terms to find
the limits. Then, as x gets very very very close to 2, we find the
limit. ## rigour Limit of a function f(x := L as x → x0
Let x0 ∈ ℝ and f
be a function defined on a punctured neighbourhood of x0 i.e. on (x0 − a, x0 + a) \ {0} = (x0 − a, x0) ∪ (x0, x0 + a)
for some a > 0 limx → x0[f(x)] = L
If and only if $|f(x) - L| < $ for all ϵ > 0 where 0 < |x − x0| < δ
for δ > 0 Continuous at a
poin := If limx → x0[f(x)]
exists and is equal to f(x0) Continuous
[[function] := If every point on the function is continous
Asymptotic limit := when a value can become arbitrarily
close to but never be a value, for instance limx → ∞ is an asymptotic
limit limit at infinity :~ ∞ is a
“singular point” on the real line, and its neighbourhood are sets of the
form (R, ∞); similarly, the
neighbourhoods of −∞ are sets of the
form (−∞, −R)
Alternate definition for limit := For any neighbourhood
U = (y0 − ϵ, y0 + ϵ)
of y0 there exists
a neighbourhood V = (x0 − δ, x0 + δ)
of x0 such that if
x ∈ V, then f(x) ∈ U
[[Intermediate value theorem]]
[[Sandwich theorem]]
Dividing each term by the highest degree in the denominator of a limit that goes to infinity
Limit of a function f(x := L as x → x0
Let x0 ∈ ℝ and f be a function defined on a punctured neighbourhood of x0 i.e. on (x0 − a, x0 + a) \ {0} = (x0 − a, x0) ∪ (x0, x0 + a) for some a > 0 limx → x0[f(x)] = L If and only if $|f(x) - L| < $ for all ϵ > 0 where 0 < |x − x0| < δ for δ > 0
f(x) = $\frac{x+1}{x+2} \to 1$ as x → ∞
Suppose limn → ∞an = a and limn → ∞bn = b
limn → ∞(an ± bn) = a ± b
limn → ∞(anbn) = ab
$\lim_{n \to \infty} \left( \frac{a_n}{b_n} \right) = \frac{a}{b}$ if b ≠ 0
tags :math:
:todo: let Sn → 𝕝 and let Tn → 𝕞 as n → ∞
for any α ∈ ℝ one has αSn → α𝕝 as n → ∞
if SnTn
When finding limits
$$\lim[\frac{f(x)}{g(x)}] = \frac{\lim[f(x)]}{lim[g(x)]}$$ provided lim [g(x)]≠0
nn ≫ n! ≫ an ≫ nk ≫ log (n)
Algebra of limits
Divide by the fastest growing term in the denominator
Subsequences, don’t try to think of a formulae for it.
A limit point is the limit of some subsequence, TODO, learn some of the properties
write out the terms of the subsequence
taking reciprorocals, generally changes the limit points
IF a_n has a limit point l /l limit point of 1/a_n
IF a_n has a limit point l = 0 the corresponding subsequence will diverge
IF a_n has a subsequence +- the corresponding subsequence will converge to 0.
These are all intuitive
Continuous at a poin := If limx → x0[f(x)]
exists and is equal to f(x0)
Continuous [[function] := If every point on the function
is continous
Asymptotic limit := when a value can become arbitrarily
close to but never be a value, for instance limx → ∞ is an asymptotic
limit
limit at infinity :~ ∞ is a “singular point” on the real line, and its neighbourhood are sets of the form (R, ∞); similarly, the neighbourhoods of −∞ are sets of the form (−∞, −R)
Alternate definition for limit := For any neighbourhood
U = (y0 − ϵ, y0 + ϵ)
of y0 there exists
a neighbourhood V = (x0 − δ, x0 + δ)
of x0 such that if
x ∈ V, then f(x) ∈ U
Intuitions/notes:
δ is dependent on ϵ, that means if you change epsilon, your value of delta should be different. Note that as it is a modulus, then naturally we begin to see that we consider points above and below L. We are making a rigorous definition of the converging phenomena Finding if a limit exists can be done with L’hopital’s rule, and differentiation has an implied limit.
limx → x0[f(x)g(x)] = limx → x0[f(x)]limx → x0[g(x)]
limx → x0[f(x) + g(x)] = limx → x0[f(x)] + limx → x0[g(x)] (BE CAREFUL OF ∞ − ∞)
$\lim_{x\rightarrow x_0}[\frac{f(x)}{g(x)}] = \frac{\lim_{x\rightarrow x_0}[f(x)]}{\lim_{x\rightarrow x_0}[g(x)]}$
Order matters when dealing with double limits
Intuition regarding cancelling “zero” terms
when finding the limit $\lim_{x\rightarrow 2}[\frac{(2x+1)(x-2)}{x-2}]$ it makes the problem much easier when we cancel x-2 from the top and bottom. BUT this is not the same function. But what we are doing is we are saying x ≠ 2 so we cancel the terms to find the limits. Then, as x gets very very very close to 2, we find the limit.
tags: :calculus_1:calculus_2:real_analysis:sequences_and_series: