6.4 Limits[29N]
We will write \(\overline{ℝ}\) for \(ℝ∪\{ ±∞\} \).
[20D]Let \(I⊂ ℝ\), \(x_ 0∈ \overline{ℝ}\) accumulation point of \(I\), \(f:I→ ℝ\) function, \(l∈\overline{ℝ}\).
The idea of limit (right or left or bilateral) is thus expressed.
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\(\lim _{x→ x_ 0} f(x) = l\) |
for every ”full” neighbourhood \(V\) of \(l\), there exists a “deleted” neighbourhood \(U\) of \(x_ 0\) such that for every \(x∈ U∩ I\), you have \(f(x)∈ V\) |
where the neighborhood \(U\) will be “right” or “left’ if the limit is “right” or “left”; it can also be said that
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\(\lim _{x→ x_ 0} f(x) = l\) |
for every ”full” neighbourhood \(V\) of \(l\), you have \(f(x)∈ V\) eventually for \(x\) tending to \(x_ 0\) |
adding that \(x{\gt}x_ 0\) if the limit is “right”, or \(x{\lt}x_ 0\) if the limit is “left”.
[0BH] Let \(I\) be a set, \(x_ 0∈ℝ\) accumulation point for \(I\), \(f:I→ ℝ\) function, \(l∈ℝ\).
Putting together all the definitions seen above, we get these definitions of limit.
In the case \(x_ 0∈ℝ\) and \(l∈ℝ\):
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\(\lim _{x→ x_ 0} f(x) = l\) |
\(∀ \varepsilon {\gt}0, ∃ 𝛿 {\gt}0, ∀ x, |x-x_ 0|{\lt}𝛿, x≠ x_ 0, x∈ I⇒|f(x)-l|{\lt}\varepsilon \) |
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\(\lim _{x→ x_ 0^+} f(x) = l\) |
\(∀ \varepsilon {\gt}0, ∃ 𝛿 {\gt}0, ∀ x, |x-x_ 0|{\lt}𝛿, x{\gt} x_ 0, x∈ I⇒|f(x)-l|{\lt}\varepsilon \) |
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\(\lim _{x→ x_ 0^-} f(x) = l\) |
\(∀ \varepsilon {\gt}0, ∃ 𝛿 {\gt}0, ∀ x, |x-x_ 0|{\lt}𝛿, x{\lt} x_ 0, x∈ I⇒|f(x)-l|{\lt}\varepsilon \) |
Be \(x_ 0∈ℝ\), \(l=±∞\).
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\(\lim _{x→ x_ 0} f(x) = ∞\) |
\(∀ z, ∃ 𝛿 {\gt}0, ∀ x, |x-x_ 0|{\lt}𝛿, x≠ x_ 0, x∈ I⇒f(x){\gt}z \) |
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\(\lim _{x→ x_ 0} f(x) = -∞\) |
\(∀ z, ∃ 𝛿 {\gt}0, ∀ x, |x-x_ 0|{\lt}𝛿, x≠ x_ 0, x∈ I⇒f(x){\lt}z \) |
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\(\lim _{x→ x_ 0^+} f(x) = ∞\) |
\(∀ z, ∃ 𝛿 {\gt}0, ∀ x, |x-x_ 0|{\lt}𝛿, x{\gt} x_ 0, x∈ I⇒f(x){\gt}z \) |
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\(\lim _{x→ x_ 0^+} f(x) = -∞\) |
\(∀ z, ∃ 𝛿 {\gt}0, ∀ x, |x-x_ 0|{\lt}𝛿, x{\gt} x_ 0, x∈ I⇒f(x){\lt}z \) |
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\(\lim _{x→ x_ 0^-} f(x) = ∞\) |
\(∀ z, ∃ 𝛿 {\gt}0, ∀ x, |x-x_ 0|{\lt}𝛿, x{\lt} x_ 0, x∈ I⇒f(x){\gt}z \) |
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\(\lim _{x→ x_ 0^-} f(x) = -∞\) |
\(∀ z, ∃ 𝛿 {\gt}0, ∀ x, |x-x_ 0|{\lt}𝛿, x{\lt} x_ 0, x∈ I⇒f(x){\lt}z \) |
Let \(l∈ℝ\), \(x_ 0=±∞\).
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\(\lim _{x→ ∞} f(x) = l\) |
\(∀ \varepsilon {\gt}0, ∃ y, ∀ x, x{\gt}y, x∈ I⇒|f(x)-l|{\lt}\varepsilon \) |
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\(\lim _{x→ -∞} f(x) = l\) |
\(∀ \varepsilon {\gt}0, ∃ y, ∀ x, x{\lt}y, x∈ I⇒|f(x)-l|{\lt}\varepsilon \) |
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\(\lim _{x→ ∞} f(x) = ∞\) |
\(∀ z, ∃ y, ∀ x, x{\gt}y, x∈ I⇒f(x){\gt}z\) |
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\(\lim _{x→ -∞} f(x) = ∞\) |
\(∀ z, ∃ y, ∀ x, x{\lt}y, x∈ I⇒f(x){\gt}z\) |
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\(\lim _{x→ ∞} f(x) = -∞\) |
\(∀ z, ∃ y, ∀ x, x{\gt}y, x∈ I⇒f(x){\lt}z\) |
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\(\lim _{x→ -∞} f(x) = -∞\) |
\(∀ z, ∃ y, ∀ x, x{\lt}y, x∈ I⇒f(x){\lt}z\) |
[0BJ]Note that if you replace \(f↦ -f\), you switch from definitions with \(l=∞\) to those of \(l=-∞\) (and vice versa). Another symmetry is achieved by switching \(x_ 0 → -x_ 0\) and the right and left neighbourhoods.