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LIMIT OF A REAL FUNCTION


f(x) = sin(x)/x ∀ x ∈ ℝ \ {0}, the limit in 0 exists, it is a regular case; g(x) = sin(1/x) ∀ x ∈ ℝ \ {0}, the limit in 0 does not exist, it is an irregular case

Limit point or cluster point or accumulation point: c ∈ = ℝ ∪ {±∞}, X ⊆ ℝ, c is an accumulation point for X if ∃ {xn}n∈ℕ | 1) xn ∈ X ∀ n ∈ ℕ, 2) xn ≠ c ∀ n ∈ ℕ, 3) limn→∞ xn = c; c ∈ ℝ can be approximated by points contained in X; X = ℝ \ {0}, c = 0 is an accumulation point for X, xn = 1/n; X = (a,b) or [a,b], the set of accumulation points of X, identified as X, is [a,b]; the set of accumulation points of an open or closed interval is always the whole closed interval; xn = c+1/n2, c=1, xn = 1+1/n2, c=2, xn = 2-1/n; X = ℕ, the only accumulation point is c = +∞, xn = n; the only accumulation point of the natural number set is +∞; X = {-1} ∪ {1} ∪ [3,+∞), X = [3,+∞) ∪ {+∞}, -1,1 are isolated point of X; X = {1/n, n ∈ ℕ}, c = 0 is the only accumulation point, xn = 1/n

What is the set of limit points for ℚ?

f: X ⊆ ℝ → ℝ, c ∈ is an accumulation point for X, then f tends to l as x tends to c, limx→c f(x) = l, where l ∈ if ∀ {xn}n∈ℕ | xn ∈ X ∀ n ∈ ℕ, xn ≠ c ∀ n ∈ ℕ, limn→∞ xn = c ⇒ limn→∞ f(xn) = l; because c is an accumulation point for X, and exists at least a sequence {xn}n ⊆ X | xn → c for n → ∞; l must not depend on the sequence {xn}n∈ℕ that converges to the accumulation point c

f(x) = sin(x)/x ∀ x ∈ ℝ \ {0}; limx→0 sin(x)/x = 1 ⇔ ∀ {xn}n∈ℕ | xn ≠ 0 and limn→∞ xn = 0 ⇒ limn→∞ f(xn) = limn→∞ sin(xn)/xn = 1, because limn→∞ sin(an)/an = 1 if limn→∞ an = 0

∄ limx→0 sin(1/x); to prove that a limit does not exist, it is necessary to find two sequences that converge at the accumulation point, in this case 0, and the corresponding limits of the images must be different from each other; xn = 1/2πn ≠ 0, limn→∞ xn = limn→∞ 1/2πn = 0; limn→∞ g(xn) = limn→∞ sin(1/xn) = limn→∞ sin(2πn) = 0; yn = 1/(π/2+2πn) ≠ 0, limn→∞ yn = limn→∞ 1/(π/2+2πn) = 0, limn→∞ g(yn) = limn→∞ sin(1/yn) = limn→∞ sin(π/2+2πn) = 1; the limit of the images is different and it depends on the chosen sequence

This definition of limit is well defined for c ∈ ℝ ∪ {±∞} and l ∈ ℝ ∪ {±∞}

f: X ⊆ ℝ → ℝ, c ∈ ℝ, l ∈ ℝ, then limx→c f(x) = l if ∀ ε > 0 ∃ δ = δ(ε) > 0 | if x ∈ X and 0 < |x-c| < δ ⇒ |f(x)-l| < ε ⇔ l-ε < f(x) < l+ε

The definition of a limit of a real function that uses the sequences is unique, instead the definition of a limit of a real function using ε and δ must be formulated in 6 different cases, c = ℝ, c = -∞, c = + ∞, l = ℝ, l = -∞, l = + ∞

l = +∞, c ∈ ℝ, limx→c f(x) = +∞ ⇔ ∀ M ∈ ℝ ∃ δ = δ(M) > 0 | if x ∈ X, 0 < |x-c| < δ ⇒ f(x) > M

The limit value does not depend on the value at the accumulation point; the function could be defined or undefined at the accumulation point

Calculating a limit we do not see what happens in the accumulation point, but we see what happens in the neighboring points that tend to the accumulation point, in fact xn → 0, xn ≠ 0

f(x) = {1, x = 0; 0, x ≠ 0}; limx→0 f(x) = 0 ⇒ f(xn) = 0 ⇒ limn→∞ f(xn) = 0

f(x) = x/|x| = {1, x > 0; -1, x < 0}; f(x) = sgn(x) ∀ x ∈ ℝ \ {0}; limx→0 f(x) ∄; xn = 1/n → 0 for n → ∞ and limn→∞ f(xn) = limn→∞ (1/n)/|1/n| = 1, yn = -1/n → 0 for n → ∞ and limn→∞ f(yn) = limn→∞ (-1/n)/|-1/n| = -1, ⇒ ∄ limx→0 f(x)

One-sided limit: {xn}n∈ℕ, c ∈ ℝ, 1) limn→∞ xn = c+ if {limn→∞ xn = c, xn ≥ c ∀ n ∈ ℕ}, 2) limn→∞ xn = c- if {limn→∞ xn = c, xn ≤ c ∀ n ∈ ℕ}; limit from right xnn→∞ → c+, limit from left xnn→∞ → c-; limit from right: c ∈ ℝ, l ∈ , limx→c+ f(x) = l if ∀ {xn}n∈ℕ | xn ∈ X, xn ≠ c, limn→∞ xn = c+, then limn→∞ f(xn) = l; limit from left: c ∈ ℝ, l ∈ , limx→c- f(x) = l if ∀ {xn}n∈ℕ | xn ∈ X, xn ≠ c, limn→∞ xn = c-, then limn→∞ f(xn) = l

limx→0+ x/|x| = 1; limx→0- x/|x| = -1; this function admits limits from right and left, but the limits from right and left are different and therefore does not admit limit

Necessary and sufficient condition for limx→c f(x) = l is limx→c- f(x) = limx→c+ f(x) = l; if the limit exists, the limits from the right and from the left exist and coincide with the value l, vice versa if the limits from the right and from the left exist and coincide, then the limit of the function exists and has value l

limx→0 1/x; xn = 1/n → 0, limn→∞ 1/xn = limn→∞ n = +∞; yn = -1/n → 0, limn→∞ 1/yn = limn→∞ (-n) = -∞

limx→0+ 1/x = +∞, {xn} | limn→∞ xn = 0+ ⇒ limn→∞ 1/xn = 1/0+ = +∞; limx→0- 1/x = -∞

If limx→±∞ f(x) = l ∈ ℝ, then the line y = l is a horizontal asymptote in ±∞

If limx→c f(x) = ±∞, c ∈ ℝ, then the line x = c is a vertical asymptote for f

Limit and algebraic operations: limx→c f(x) = l, limx→c g(x) = m, m,l ∈ ℝ; limx→c (f(x)+g(x)) = l+m; limx→c (f(x)-g(x)) = l-m; limx→c f(x)⋅g(x) = l⋅m; limx→c f(x)/g(x) = l/m, m ≠ 0; limx→c f(x)g(x) = lm, l > 0; limx→c |f(x)| = |l|; if l,m ∈ these rules are useful for the determinate forms, in the case of an indeterminate form this is useless

Limit of compound function: f: X ⊆ ℝ → Y, g: Y → ℝ, c is an accumulation point for X, 1) limx→c f(x) = y0, 2) limy→y0 g(y) = l, 3) ∃ δ > 0 | 0 < |x-c| < δ, x ∈ X ⇒ f(x) ≠ y0, then limx→c (g∘f)(x) = limx→c g(f(x)) = l

The limit of a compound function is the limit of the external function with accumulation point given by the limit of the internal function

limx→+∞ sin(1/x2); h(x) = sin(1/x2) = g(f(x)) = {f(x) = 1/x2; g(y) = sin(y)}; limx→+∞ 1/x2 = 0, limx→+∞ sin(1/x2) = limy→0 sin(y) = 0

f(x) = 0 ∀ x ∈ ℝ; g(y) = {1, x ≠ 0; 0, x = 0}; limx→0 g(f(x)) = limx→0 g(0) = 0; limy→0 g(y) = 1; the condition f(x) ≠ y0 is not met

Order of limits for real functions: f,g: X ⊆ ℝ → ℝ, c is the accumulation point for X, f(x) ≤ g(x) ∀ x ∈ X, limx→c f(x) = l, limx→c g(x) = m, then l ≤ m

Squeeze theorem or theorem of carabinieri for real functions: f,g,h: X → ℝ, f(x) ≤ h(x) ≤ g(x) ∀ x ∈ X, limx→c f(x) = limx→c g(x) = l, then limx→c h(x) = l

Theorem of permanence of the sign for real functions: f: X ⊆ ℝ → ℝ, f(x) ≥ 0 ∀ x ∈ X, limx→c f(x) = l, then l ≥ 0

Generalizing the permanence of the sign, if f(x) ∈ [α,β] then l ∈ [α,β]

The theorem of permanence of the sign is true for not strict inequalities which are ≥ and ≤, but it is false for strict inequalities which are > and <; f(x) = 1/x > 0 ∀ x ∈ (0,+∞), limx→+∞ 1/x = 0

limx→0 sin(x)/x = 1; x > 0 ⇒ sin(x) > 0, 0 ≤ sin(x) ≤ x ≤ tan(x) ≤ sin(x)/cos(x); sin(x)/sin(x) ≤ x/sin(x) ≤ (sin(x)/cos(x))(1/sin(x)); 1 ≤ x/sin(x) ≤ 1/cos(x); cos(x) ≤ sin(x)/x ≤ 1, for x→0+ cos(x) goes to 1, so for the squeeze theorem or theorem of carabinieri limx→0+ sin(x)/x = 1